Q.What is the total number of orbitals associated with the principal quantum number n = 3?

Showing the 12 most recent of 16 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.Match the following

   [!FORMULA] $\begin{array}{|c|c|} \hline \text{List-1 (Element)} & \text{List-2 (Block)} \\ \hline \text{A Hf} & \text{I s-block} \\ \text{B Ra} & \text{II p-Block} \\ \text{C Am} & \text{III d-Block} \\ \text{D At} & \text{IV f-Block} \\ \hline \end{array}$

   (A) A – IV, B – III, C – I, D – II (B) A – II, B – III, C – IV, D – I (C) A – III, B – IV, C – I, D – II (D) A – III, B – I, C – IV, D – II

   ›Reveal solutionSolution

   The problem asks to match each element (Hf, Ra, Am, At) to its block in the periodic table (s, p, d, f). The correct matches are: Hf (d-block), Ra (s-block), Am (f-block), At (p-block), so the answer is option (D).

   The periodic table is organized into blocks based on which subshell (s, p, d, or f) the element’s outermost electrons occupy. This is determined by the element’s electron configuration, which follows the Aufbau principle. The key is to know the position of each element in the periodic table, or to deduce its block from its group and period.

   Let’s match each element step by step:

   1. Hf (Hafnium)

      • Hafnium is element 72. It lies in period 6, group 4.
      • Group 4 elements are transition metals, which are always in the d-block.
      • Its electron configuration ends in $5d^2 6s^2$, confirming it’s a d-block element.
      • So A matches III (d-block).

   2. Ra (Radium)

      • Radium is element 88, in period 7, group 2.
      • Group 2 elements are alkaline earth metals, which are in the s-block (the last electron enters the s subshell).
      • Configuration: $[Rn] 7s^2$.
      • So B matches I (s-block).

   3. Am (Americium)

      • Americium is element 95, an actinide.
      • Actinides are part of the f-block (the 5f subshell is being filled).
      • Configuration: $[Rn] 5f^7 7s^2$.
      • So C matches IV (f-block).

   4. At (Astatine)

      • Astatine is element 85, in period 6, group 17 (halogens).
      • Group 17 elements are in the p-block (the last electron enters a p subshell).
      • Configuration: $[Xe] 4f^{14} 5d^{10} 6p^5$.
      • So D matches II (p-block).

   Thus the correct pairing is: A–III, B–I, C–IV, D–II. Looking at the options, this corresponds to option (D).

   Watch out

   A common mistake is to confuse Radium (Ra) with a d-block element because it’s a metal, but it’s actually an s-block alkaline earth metal. Similarly, Americium (Am) is often misclassified as a d-block element, but it’s an f-block actinide.

   Tip

   Remember the block boundaries: groups 1-2 are s-block, groups 3-12 are d-block, groups 13-18 are p-block, and the two rows below (lanthanides and actinides) are f-block. This makes matching quick without needing full configurations.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.The energy of orbit X of Li$^{2+}$ (Z = 3) is $-2.18 \times 10^{-18}$ J. What is the radius of the same orbit (in $\text{\AA}$)? (A) 2.116 (B) 2.105 (C) 1.587 (D) 2.645

   ›Reveal solutionSolution

   The key idea is to use the Bohr model for hydrogen-like ions: energy depends on $n^2/Z^2$ and radius on $n^2/Z$. Given the energy of Li$^{2+}$, we first find the principal quantum number $n$, then compute the radius. The radius comes out to $1.587 \, \text{Å}$, so the correct option is (C).

   ---

   We are dealing with a hydrogen-like ion (Li$^{2+}$ has only one electron, so the Bohr model applies exactly). In the Bohr model, the energy of an electron in orbit $n$ for a nucleus of charge $Z$ is:

   $E_n = - \frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2}$

   But here the energy is given in joules. The ground-state energy of hydrogen ($Z=1$, $n=1$) is $-13.6 \, \text{eV} = -2.18 \times 10^{-18} \, \text{J}$. So we can write:

   $E_n = - \frac{Z^2}{n^2} \times (2.18 \times 10^{-18} \, \text{J})$

   For Li$^{2+}$, $Z = 3$, and we are told $E_n = -2.18 \times 10^{-18} \, \text{J}$. That means:

   $- \frac{3^2}{n^2} \times (2.18 \times 10^{-18}) = -2.18 \times 10^{-18}$

   Cancelling the common factor:

   $\frac{9}{n^2} = 1 \quad \Rightarrow \quad n^2 = 9 \quad \Rightarrow \quad n = 3$

   So the electron is in the third Bohr orbit.

   Now, the radius of the $n$th orbit for a hydrogen-like ion is:

   $r_n = \frac{n^2}{Z} \, a_0$

   where $a_0 = 0.529 \, \text{Å}$ is the Bohr radius.

   Plug in $n=3$, $Z=3$:

   $r_3 = \frac{3^2}{3} \times 0.529 \, \text{Å} = \frac{9}{3} \times 0.529 = 3 \times 0.529 = 1.587 \, \text{Å}$

   Thus the radius is $1.587 \, \text{Å}$.

   Watch out

   A common mistake is to forget that the given energy ($-2.18 \times 10^{-18}$ J) is exactly the hydrogen ground-state energy. Students sometimes plug numbers into formulas without noticing that this immediately forces $n=3$ for Li$^{2+}$.

   Tip

   Notice that the energy given is numerically the same as the hydrogen ground-state energy. Since energy scales as $Z^2/n^2$, for $Z=3$ we need $n=3$ to get the same value. This shortcut saves time.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

3. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.The atomic numbers of four elements A, D, E, G are 4, 7, 8, 12 respectively. The decreasing order of electronegativity of these elements is (A) E, D, A, G (B) E, A, D, G (C) D, G, E, A (D) G, A, D, E

   ›Reveal solutionSolution

   Electronegativity increases across a period and decreases down a group. Using the periodic table positions of elements with atomic numbers 4 (Be), 7 (N), 8 (O), and 12 (Mg), the decreasing order is E, D, A, G, which corresponds to option (A).

   The key to ordering electronegativity is understanding the periodic trends. Electronegativity — the ability of an atom to attract shared electrons in a bond — increases as you move from left to right across a period (because nuclear charge increases while atomic radius decreases) and decreases as you move down a group (because the valence shell gets farther from the nucleus, reducing attraction). So the first step is to identify each element from its atomic number and place it in the periodic table.

   1. Identify the elements.

      Atomic number 4 is beryllium (Be, group 2, period 2).

      Atomic number 7 is nitrogen (N, group 15, period 2).

      Atomic number 8 is oxygen (O, group 16, period 2).

      Atomic number 12 is magnesium (Mg, group 2, period 3).

      So A = Be, D = N, E = O, G = Mg.

   2. Compare elements in the same period.

      Be, N, and O are all in period 2. Across a period, electronegativity increases from left to right. The order in period 2 (lowest to highest) is: Be < B < C < N < O < F. So among these three:

      Be (A) has the lowest, then N (D), then O (E) has the highest.

      So far: E > D > A.

   3. Place the element from the next period.

      Mg (G) is directly below Be in group 2. Down a group, electronegativity decreases. So Mg has a lower electronegativity than Be. Since Be is already the lowest among the period-2 elements, Mg is even lower.

      So the full decreasing order is: O (E) > N (D) > Be (A) > Mg (G).

   Watch out

   A common mistake is to forget that Mg is below Be in group 2 and assume it has a higher electronegativity because it has more protons. But the increase in atomic radius and additional shielding dominate, making Mg less electronegative than Be.

   4. Match with the options. The sequence E, D, A, G corresponds exactly to option (A).

   ✓Final answer

   The correct option is (A).

4. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.A metal was irradiated separately with radiation of wavelengths

   (a) 400 nm

   (b) 500 nm and

   (c) 600 nm respectively. Identify the wavelength(s) corresponding to which electrons are emitted from the surface of that metal (Work function of metal = 2.25 eV, $h = 6.6 \times 10^{-34}$ Js, $c = 3 \times 10^8$ ms$^{-1}$, $1$eV $= 1.6 \times 10^{-19}$ J) (A)

   (a) only (B)

   (c) only (C) (a),

   (b) &

   (c) (D)

   (a) &

   (b) only

   ›Reveal solutionSolution

   The photoelectric effect requires the incident photon energy to exceed the metal’s work function. Only wavelengths shorter than the threshold wavelength cause emission. Here, only 400 nm and 500 nm satisfy this, so the answer is (D).

   The core idea is the photoelectric effect: an electron is ejected from a metal surface only if the energy of an incident photon is at least equal to the work function $\phi$ of the metal. The photon energy is $E = \frac{hc}{\lambda}$, so for a given work function, there is a maximum wavelength (threshold wavelength $\lambda_0$) beyond which no emission occurs. Any wavelength shorter than $\lambda_0$ will cause emission; any longer will not.

   We are given $\phi = 2.25$ eV, $h = 6.6 \times 10^{-34}$ Js, $c = 3 \times 10^8$ m/s, and $1$ eV $= 1.6 \times 10^{-19}$ J. The three wavelengths are 400 nm, 500 nm, and 600 nm. We need to compare each photon’s energy to $\phi$.

   1. Find the threshold wavelength $\lambda_0$. Set the photon energy equal to the work function:

   $\frac{hc}{\lambda_0} = \phi$

   Convert $\phi$ to joules:

   $\phi = 2.25 \times 1.6 \times 10^{-19} = 3.6 \times 10^{-19} \text{ J}$

   Then

   $\lambda_0 = \frac{hc}{\phi} = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{3.6 \times 10^{-19}}$

   Calculate:

   $\lambda_0 = \frac{19.8 \times 10^{-26}}{3.6 \times 10^{-19}} = 5.5 \times 10^{-7} \text{ m} = 550 \text{ nm}$

   So the threshold wavelength is 550 nm. Any wavelength shorter than 550 nm will eject electrons; any longer will not.

   2. Compare each given wavelength to $\lambda_0$.
      • 400 nm < 550 nm → photon energy > work function → electrons emitted.
      • 500 nm < 550 nm → photon energy > work function → electrons emitted.
      • 600 nm > 550 nm → photon energy < work function → no emission.

   Watch out

   A common mistake is to forget unit conversion. The work function is given in eV, but $h$ and $c$ are in SI units. Always convert $\phi$ to joules before plugging into $hc/\lambda$, or convert $hc$ to eV·nm for a shortcut. Here, $hc \approx 1240$ eV·nm, so $\lambda_0 = 1240 / 2.25 \approx 551$ nm — same result.

   3. Identify the correct option. Emission occurs for 400 nm and 500 nm only. That matches option (D): (a) & (b) only.

   ✓Final answer

   The correct option is (D).

5. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.In which of the following options, elements are correctly arranged in the increasing order of their atomic radius? (A) Si < P < Na < N < F (B) Na < Si < P < N < F (C) F < N < P < Si < Na (D) N < F < Si < P < Na

   ›Reveal solutionSolution

   Atomic radius increases down a group and decreases across a period; by applying periodic trends, the correct increasing order is F < N < P < Si < Na, which corresponds to option (C).

   The key concept here is periodic trends in atomic radius. As you move left to right across a period, the nuclear charge increases, pulling electrons closer and making the atom smaller. As you move down a group, new electron shells are added, making the atom larger. To compare radii of different elements, you need to place them in the periodic table and apply these two rules together.

   Let’s work through the options step by step.

   1. Identify the positions of the elements

      • Na (sodium): Group 1, Period 3
      • Si (silicon): Group 14, Period 3
      • P (phosphorus): Group 15, Period 3
      • N (nitrogen): Group 15, Period 2
      • F (fluorine): Group 17, Period 2

   2. Compare within the same period

      In Period 3: Na, Si, P. Atomic radius decreases from left to right:

   $\text{Na} > \text{Si} > \text{P}$

   In Period 2: N and F. Atomic radius decreases left to right:

   $\text{N} > \text{F}$

   3. Compare across periods

      Elements in Period 2 (N, F) are smaller than any element in Period 3 because they have fewer electron shells. So the smallest atoms overall are F and N.

      Between N and F: F is smaller (further right in Period 2).

      So the smallest is F, then N.

   4. Order the Period 3 elements

      After N, the next smallest is P (Period 3, far right), then Si (Period 3, middle), and the largest is Na (Period 3, far left).

      So the full increasing order is:

   $\text{F} < \text{N} < \text{P} < \text{Si} < \text{Na}$

   5. Match with the options
      • (A) Si < P < Na < N < F → wrong (N and F are smallest, not largest)
      • (B) Na < Si < P < N < F → wrong (Na is largest, not smallest)
      • (C) F < N < P < Si < Na → correct
      • (D) N < F < Si < P < Na → wrong (F is smaller than N)

   Tip

   A common mistake is forgetting that elements in higher periods are always larger than those in lower periods, even if they are far to the right. For example, P (Period 3) is larger than N (Period 2), even though P is to the right of Si.

   Watch out

   Do not compare atomic radii by atomic number alone — periodic trends depend on both shell number and nuclear charge. For instance, N (Z=7) is smaller than Si (Z=14) because N has fewer shells, not because of atomic number.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

6. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.According to Bohr’s theory of hydrogen atom the approximate angular momentum of electron in H atom in the ground state is ($h = 6.62 \times 10^{-34}$ Js) (A) $1.05 \times 10^{-34}$ J (B) $1.05 \times 10^{-34}$ Js (C) $6.62 \times 10^{+34}$ J (D) $6.62 \times 10^{-34}$ Jm

   ›Reveal solutionSolution

   According to Bohr's theory, the angular momentum of an electron in a stable orbit is quantized. For the ground state ($n=1$) of the hydrogen atom, the angular momentum is $L = \frac{h}{2\pi}$, which evaluates to approximately $\boxed{1.05 \times 10^{-34} \text{ Js}}$.

   The Bohr model of the hydrogen atom introduced several revolutionary postulates to explain the stability of atoms and the discrete nature of atomic spectra. One of its key postulates was the quantization of angular momentum. This means that an electron can only revolve in certain specific orbits, and in these orbits, its angular momentum must be an integral multiple of a fundamental constant.

   This postulate directly addresses a major problem with classical physics: according to classical electromagnetism, an electron orbiting a nucleus should continuously radiate energy and spiral into the nucleus. Bohr proposed that electrons in these "stationary orbits" do not radiate energy, and the condition for these stable orbits is that the electron's angular momentum is quantized.

   1. Recall Bohr's Quantization Condition for Angular Momentum:

      According to Bohr's theory, the angular momentum ($L$) of an electron in a stationary orbit is quantized. It can only take on discrete values given by the formula:

      $L = n \frac{h}{2\pi}$

      where:

      • $L$ is the angular momentum of the electron.
      • $n$ is the principal quantum number, which can be $1, 2, 3, \dots$ (corresponding to the first, second, third, etc., allowed orbits).
      • $h$ is Planck's constant.

   2. Identify the Principal Quantum Number for the Ground State:

      The question asks for the angular momentum of the electron in the ground state of the hydrogen atom. The ground state corresponds to the lowest energy level, which means the principal quantum number $n=1$.

   3. Substitute Values and Calculate:

      We are given Planck's constant $h = 6.62 \times 10^{-34}$ Js.

      For the ground state, $n=1$.

      Substitute these values into the formula:

      $L = 1 \times \frac{6.62 \times 10^{-34} \text{ Js}}{2\pi}$

      Using $\pi \approx 3.14159$:

      $L = \frac{6.62 \times 10^{-34}}{2 \times 3.14159} \text{ Js}$

      $L = \frac{6.62 \times 10^{-34}}{6.28318} \text{ Js}$

      $L \approx 1.0536 \times 10^{-34} \text{ Js}$

      Rounding to two decimal places, we get:

      $L \approx 1.05 \times 10^{-34} \text{ Js}$

   4. Compare with Options:

      Let's examine the given options:

      (A) $1.05 \times 10^{-34}$ J (Incorrect unit; angular momentum is in Js)

      (B) $1.05 \times 10^{-34}$ Js (Correct value and unit)

      (C) $6.62 \times 10^{+34}$ J (Incorrect magnitude and unit)

      (D) $6.62 \times 10^{-34}$ Jm (Incorrect magnitude and unit)

      The calculated value matches option (B).

   Important

   The quantity $\frac{h}{2\pi}$ is often denoted by $\hbar$ (read as "h-bar" or "reduced Planck's constant"). So, Bohr's quantization condition can be written as $L = n\hbar$. The angular momentum in the ground state ($n=1$) is simply $\hbar$.

   ✓Final answer

   The approximate angular momentum of the electron in the ground state of the hydrogen atom is $\boxed{1.05 \times 10^{-34} \text{ Js}}$.

7. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.The wavelength of second line of Balmer series of hydrogen atom is $\lambda$ nm. What is the wavelength of first line of Lyman series of He$^+$ ion (in nm)? (A) $\dfrac{\lambda}{16}$ (B) $\dfrac{16}{\lambda}$ (C) $\dfrac{16}{3\lambda}$ (D) $\dfrac{3\lambda}{16}$

   ›Reveal solutionSolution

   Balmer 2nd line of H gives $\dfrac{1}{\lambda}=\dfrac{3R}{16}$; the first Lyman line of $\text{He}^+$ then has wavelength $\dfrac{\lambda}{16}$.

   Using the Rydberg formula $\dfrac{1}{\lambda}=RZ^{2}\left(\dfrac{1}{n_1^{2}}-\dfrac{1}{n_2^{2}}\right)$.

   Second line of the Balmer series of H ($Z=1$, $n_1=2$, and the second line is $n_2=4$):

   $\frac{1}{\lambda}=R(1)^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=R\left(\frac{1}{4}-\frac{1}{16}\right)=R\cdot\frac{3}{16}.$

   So $R=\dfrac{16}{3\lambda}$.

   First line of the Lyman series of $\text{He}^+$ ($Z=2$, $n_1=1$, first line is $n_2=2$):

   $\frac{1}{\lambda'}=R(2)^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=4R\cdot\frac{3}{4}=3R.$

   Therefore

   $\lambda'=\frac{1}{3R}=\frac{1}{3\cdot\frac{16}{3\lambda}}=\frac{\lambda}{16}.$

   ✓Final answer

   $\lambda'=\dfrac{\lambda}{16}$ nm — option (A).

8. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.In which of the following options, the elements are correctly arranged with respect to their negative electron gain enthalpies? (A) P > S > Cl > F (B) S > P > F > Cl (C) Cl > F > S > P (D) F > Cl > P > S

   ›Reveal solutionSolution

   The key idea is that electron gain enthalpy (EA) becomes more negative across a period, but anomalies occur due to atomic size and electron repulsion. The correct order is Cl > F > S > P, which corresponds to option (C).

   The concept here is electron gain enthalpy — the energy change when an atom gains an electron to form a negative ion. A more negative value means a stronger tendency to accept an electron. While we might expect a simple trend (more negative across a period, less negative down a group), exceptions arise because of small atomic size (leading to electron-electron repulsion) and half-filled/full-filled stability.

   Let’s reason through the given elements: P, S, Cl, F.

   1. Chlorine vs. Fluorine

      Chlorine has a more negative electron gain enthalpy than fluorine. Why? Fluorine is very small; when it gains an electron, the incoming electron experiences strong repulsion from the electrons already present. Chlorine is larger, so the added electron is less repelled and more easily accommodated. Thus, Cl > F in negativity.

   2. Sulfur vs. Phosphorus

      Sulfur is to the right of phosphorus in the same period. Across a period, nuclear charge increases, making electron gain more exothermic. So sulfur has a more negative EA than phosphorus: S > P.

   3. Comparing the two groups

      Now we need to place all four in order. Chlorine is the most negative overall. Fluorine is less negative than chlorine but still more negative than sulfur? Actually, fluorine’s EA is less negative than chlorine’s but still more negative than sulfur’s. However, sulfur’s EA is more negative than phosphorus’s. So the descending order of negativity is:

      Cl > F > S > P.

   4. Checking the options

      • (A) P > S > Cl > F — wrong, reverses the trend.
      • (B) S > P > F > Cl — wrong, places Cl last.
      • (C) Cl > F > S > P — matches our reasoning.
      • (D) F > Cl > P > S — wrong, overestimates F and misorders S and P.

   Watch out

   A common mistake is to assume fluorine has the most negative EA because it is the most electronegative. But EA and electronegativity are different; the small size of fluorine causes extra repulsion, lowering its EA relative to chlorine.

   Tip

   Remember the mnemonic: “Chlorine wins the electron contest, fluorine is second despite being small, sulfur beats phosphorus, and phosphorus is last.”

   ✓Final answer

   The correct option is (C).

   ANSWER: C

9. TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
   Q.The radius of third orbit of $\mathrm{Li}^{2+}$ ion is $x$ nm. The radius of fourth orbit of $\mathrm{He}^{+}$ ion (in nm) is (A) $\dfrac{3}{8}x$ (B) $\dfrac{8}{3}x$ (C) $\dfrac{2}{3}x$ (D) $\dfrac{3}{2}x$

   ›Reveal solutionSolution

   The radius of a hydrogen-like orbit scales as $n^2/Z$. Using the given radius $x$ for $\mathrm{Li}^{2+}$ (third orbit), the radius of the fourth orbit of $\mathrm{He}^{+}$ comes out to $\frac{8}{3}x$.

   The key idea here is the Bohr radius formula for hydrogen-like ions. For any one-electron species (nucleus of charge $+Ze$ with a single electron), the radius of the $n$th orbit is given by:

   $r_n = \frac{n^2}{Z} a_0$

   where $a_0$ is the Bohr radius (a constant, about $0.529 \, \text{Å}$). The formula comes from balancing the Coulomb attraction with the centripetal force and quantising angular momentum — but for this problem, all you need is the proportionality: $r_n \propto n^2/Z$.

   Let’s work through it step by step.

   1. Write the radius for $\mathrm{Li}^{2+}$. For $\mathrm{Li}^{2+}$, the atomic number $Z = 3$ (lithium has 3 protons). The third orbit means $n = 3$. So:

   $r_{\mathrm{Li}^{2+}, n=3} = \frac{3^2}{3} a_0 = \frac{9}{3} a_0 = 3 a_0$

   This is given as $x$ nm. So $x = 3 a_0$.

   2. Write the radius for $\mathrm{He}^{+}$. For $\mathrm{He}^{+}$, $Z = 2$ (helium has 2 protons). The fourth orbit means $n = 4$. So:

   $r_{\mathrm{He}^{+}, n=4} = \frac{4^2}{2} a_0 = \frac{16}{2} a_0 = 8 a_0$

   3. Express the $\mathrm{He}^{+}$ radius in terms of $x$. Since $x = 3 a_0$, we have $a_0 = x/3$. Substitute:

   $r_{\mathrm{He}^{+}, n=4} = 8 \cdot \frac{x}{3} = \frac{8}{3} x$

   Watch out

   A common mistake is to forget that $Z$ appears in the denominator. Students sometimes write $r_n \propto n^2 Z$ by misremembering the formula. Always check: a higher nuclear charge pulls the electron in tighter, so radius decreases with $Z$ — hence $n^2/Z$.

   Tip

   You can also solve this without ever writing $a_0$ explicitly. Just take the ratio:

   $\frac{r_{\mathrm{He}^{+},4}}{r_{\mathrm{Li}^{2+},3}} = \frac{(4^2/2)}{(3^2/3)} = \frac{16/2}{9/3} = \frac{8}{3}$

   So $r_{\mathrm{He}^{+},4} = \frac{8}{3} x$ directly. This ratio method is faster and avoids constants.

   ✓Final answer

   The radius of the fourth orbit of $\mathrm{He}^{+}$ is $\boxed{\frac{8}{3}x}$, which corresponds to option (B).

10. TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
    Q.Number of lone pairs of electrons present in Xenon atom of XeF$_2$, XeF$_4$ and XeF$_6$ respectively are (A) 1, 2, 0 (B) 3, 2, 0 (C) 3, 1, 1 (D) 3, 2, 1

    ›Reveal solutionSolution

    The number of lone pairs on the central xenon atom is determined by its total valence electrons minus the number of bonds formed. For XeF$_2$, XeF$_4$, and XeF$_6$, the lone pairs are 3, 2, and 1 respectively, which corresponds to option (D).

    The key idea is that xenon, being a noble gas, has 8 valence electrons. In each of these fluorides, it forms single bonds with fluorine atoms, using one electron per bond. The remaining electrons stay as lone pairs on xenon. The geometry around xenon is determined by the total number of electron pairs (bonding + lone), following the VSEPR theory.

    Let’s work through each molecule step by step.

    1. XeF$_2$

       Xenon has 8 valence electrons. It forms 2 bonds with fluorine atoms, using 2 electrons. That leaves $8 - 2 = 6$ electrons, which form 3 lone pairs.

       The total electron pairs around Xe = 2 bonding + 3 lone = 5 pairs, giving a trigonal bipyramidal electron geometry. The lone pairs occupy the equatorial positions, making the molecule linear.

    2. XeF$_4$

       Xenon forms 4 bonds, using 4 electrons. Remaining electrons: $8 - 4 = 4$, which form 2 lone pairs.

       Total electron pairs = 4 bonding + 2 lone = 6 pairs, giving an octahedral electron geometry. The lone pairs occupy opposite positions (trans), making the molecule square planar.

    3. XeF$_6$

       Xenon forms 6 bonds, using 6 electrons. Remaining electrons: $8 - 6 = 2$, which form 1 lone pair.

       Total electron pairs = 6 bonding + 1 lone = 7 pairs. The geometry is distorted octahedral (or pentagonal bipyramidal in electron pair arrangement), with the lone pair causing distortion.

    Watch out

    A common mistake is to forget that xenon can expand its octet. It uses d-orbitals to accommodate more than 8 electrons in bonding, but the lone pair count is still based on the original 8 valence electrons minus the number of bonds.

    Thus, the lone pairs are 3, 2, and 1 respectively.

    ✓Final answer

    The correct option is (D) 3, 2, 1.

11. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
    Q.What is the approximate angular momentum (in J s) of electron in hydrogen atom in its ground state? ($h = 6.625 \times 10^{-34}$ J s) (A) $2110 \times 10^{-37}$ (B) $2110 \times 10^{-36}$ (C) $1055 \times 10^{-34}$ (D) $1055 \times 10^{-36}$

    ›Reveal solutionSolution

    The angular momentum of an electron in the ground state of hydrogen is quantized as $\frac{h}{2\pi}$, which evaluates to approximately $1.055 \times 10^{-34} \, \text{J s}$, matching option (C).

    The key idea here is Bohr’s quantization condition for angular momentum. In the Bohr model of the hydrogen atom, the electron’s angular momentum in a stationary state is an integer multiple of $\frac{h}{2\pi}$. For the ground state ($n = 1$), this is simply $\frac{h}{2\pi}$. The problem tests whether you recall this fundamental quantum condition and can compute the numerical value correctly, paying attention to powers of ten.

    1. Recall Bohr’s postulate Bohr proposed that the angular momentum $L$ of an electron in a hydrogen atom is quantized as:

    $L = n \frac{h}{2\pi}$

    where $n$ is the principal quantum number (1, 2, 3, …) and $h$ is Planck’s constant.

    2. Apply to the ground state For the ground state, $n = 1$, so:

    $L = \frac{h}{2\pi}$

    3. Substitute the given value of $h$ $h = 6.625 \times 10^{-34} \, \text{J s}$. Thus:

    $L = \frac{6.625 \times 10^{-34}}{2\pi}$

    4. Compute numerically $2\pi \approx 6.2832$. So:

    $L \approx \frac{6.625}{6.2832} \times 10^{-34} \approx 1.0545 \times 10^{-34} \, \text{J s}$

    This is approximately $1.055 \times 10^{-34} \, \text{J s}$.

    5. Match with the options Option (C) is $1055 \times 10^{-34}$, which is $1.055 \times 10^{-34}$ when written in proper scientific notation. The other options have incorrect powers of ten or wrong coefficients.

    Watch out

    A common mistake is to forget the factor $2\pi$ and simply take $L = h$, which would give $6.625 \times 10^{-34}$ — not among the options, but close to $1055 \times 10^{-34}$ if you misplace the decimal. Always remember the $2\pi$ denominator.

    Tip

    The quantity $\frac{h}{2\pi}$ appears so often in quantum mechanics that it has its own symbol: $\hbar$ (“h-bar”). So the ground-state angular momentum is simply $\hbar$.

    ✓Final answer

    The correct option is (C).

    ANSWER: C

12. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
    Q.The correct order of ionic radii for the given species is (A) $\mathrm{Na}^+ > \mathrm{Al}^{3+} > \mathrm{Mg}^{2+} > \mathrm{K}^+$ (B) $\mathrm{K}^+ > \mathrm{Na}^+ > \mathrm{Mg}^{2+} > \mathrm{Al}^{3+}$ (C) $\mathrm{K}^+ > \mathrm{Na}^+ > \mathrm{Al}^{3+} > \mathrm{Mg}^{2+}$ (D) $\mathrm{Al}^{3+} > \mathrm{Mg}^{2+} > \mathrm{K}^+ > \mathrm{Na}^+$

    ›Reveal solutionSolution

    Ionic radii are primarily determined by the number of electron shells and, for isoelectronic species, by the effective nuclear charge. $\mathrm{K}^+$ has more electron shells, making it the largest, while among the isoelectronic $\mathrm{Na}^+, \mathrm{Mg}^{2+}, \mathrm{Al}^{3+}$, increasing nuclear charge leads to decreasing size. The correct order is $\boxed{\mathrm{K}^+ > \mathrm{Na}^+ > \mathrm{Mg}^{2+} > \mathrm{Al}^{3+}}$.

    The size of an ion, its ionic radius, is a fundamental property that dictates many chemical and physical characteristics. When comparing ionic radii, two primary factors are at play: the number of electron shells and the effective nuclear charge.

    1. Number of Electron Shells: This is the most significant factor. Ions with more electron shells will inherently be larger than ions with fewer electron shells, assuming similar nuclear charges. This is because the outermost electrons are further from the nucleus.
    2. Effective Nuclear Charge ($Z_{eff}$): For ions that have the same number of electron shells (i.e., they are isoelectronic), the ionic radius is inversely proportional to the effective nuclear charge. A higher effective nuclear charge means the nucleus pulls the electrons more strongly, drawing the electron cloud closer and resulting in a smaller ionic radius.

    Let's apply these concepts to the given species.

    1. Determine the electron configuration and number of electrons for each ion.

       • $\mathrm{Na}^+$: Sodium (Na) is in Group 1, Period 3. It has 11 protons and 11 electrons. $\mathrm{Na}^+$ loses one electron, so it has 10 electrons. Its electron configuration is $1s^2 2s^2 2p^6$, which is the same as Neon (Ne). It has 2 electron shells.
       • $\mathrm{Al}^{3+}$: Aluminium (Al) is in Group 13, Period 3. It has 13 protons and 13 electrons. $\mathrm{Al}^{3+}$ loses three electrons, so it has 10 electrons. Its electron configuration is $1s^2 2s^2 2p^6$, also like Neon. It has 2 electron shells.
       • $\mathrm{Mg}^{2+}$: Magnesium (Mg) is in Group 2, Period 3. It has 12 protons and 12 electrons. $\mathrm{Mg}^{2+}$ loses two electrons, so it has 10 electrons. Its electron configuration is $1s^2 2s^2 2p^6$, also like Neon. It has 2 electron shells.
       • $\mathrm{K}^+$: Potassium (K) is in Group 1, Period 4. It has 19 protons and 19 electrons. $\mathrm{K}^+$ loses one electron, so it has 18 electrons. Its electron configuration is $1s^2 2s^2 2p^6 3s^2 3p^6$, which is the same as Argon (Ar). It has 3 electron shells.

    2. Group the ions based on the number of electron shells.

       • $\mathrm{Na}^+, \mathrm{Al}^{3+}, \mathrm{Mg}^{2+}$ all have 2 electron shells (10 electrons). These are isoelectronic species.
       • $\mathrm{K}^+$ has 3 electron shells (18 electrons).

    3. Compare ions with different numbers of electron shells.

       Since $\mathrm{K}^+$ has 3 electron shells, while $\mathrm{Na}^+, \mathrm{Al}^{3+}, \mathrm{Mg}^{2+}$ all have 2 electron shells, $\mathrm{K}^+$ will be significantly larger than any of the other three ions.

       So, $\mathrm{K}^+$ is the largest.

    4. Compare the isoelectronic species ($\mathrm{Na}^+, \mathrm{Mg}^{2+}, \mathrm{Al}^{3+}$).

       For isoelectronic species, the ionic radius decreases as the nuclear charge (number of protons) increases.

       • $\mathrm{Na}^+$ has 11 protons.
       • $\mathrm{Mg}^{2+}$ has 12 protons.
       • $\mathrm{Al}^{3+}$ has 13 protons.

       As the number of protons increases, the positive charge in the nucleus increases, pulling the 10 electrons more strongly towards the nucleus. This results in a smaller ionic radius.

       Therefore, the order of ionic radii for these isoelectronic species is:

       $\mathrm{Na}^+ > \mathrm{Mg}^{2+} > \mathrm{Al}^{3+}$

       Watch out

       A common mistake is to assume that a higher positive charge always means a larger ion. For isoelectronic species, the opposite is true: a higher positive charge (more protons) means a stronger pull on the same number of electrons, leading to a smaller ionic radius.

    5. Combine the comparisons to get the overall order.

       We established that $\mathrm{K}^+$ is the largest, and then we ordered the isoelectronic species.

       Combining these, the correct order of ionic radii is:

       $\mathrm{K}^+ > \mathrm{Na}^+ > \mathrm{Mg}^{2+} > \mathrm{Al}^{3+}$

       Let's check this against the given options:

       (A) $\mathrm{Na}^+ > \mathrm{Al}^{3+} > \mathrm{Mg}^{2+} > \mathrm{K}^+$ (Incorrect)

       (B) $\mathrm{K}^+ > \mathrm{Na}^+ > \mathrm{Mg}^{2+} > \mathrm{Al}^{3+}$ (Correct)

       (C) $\mathrm{K}^+ > \mathrm{Na}^+ > \mathrm{Al}^{3+} > \mathrm{Mg}^{2+}$ (Incorrect, $\mathrm{Mg}^{2+}$ is larger than $\mathrm{Al}^{3+}$)

       (D) $\mathrm{Al}^{3+} > \mathrm{Mg}^{2+} > \mathrm{K}^+ > \mathrm{Na}^+$ (Incorrect)

    ✓Final answer

    The correct order of ionic radii for the given species is $\boxed{\mathrm{K}^+ > \mathrm{Na}^+ > \mathrm{Mg}^{2+} > \mathrm{Al}^{3+}}$, which corresponds to option (B).