The electron energy in hydrogen atom is given by $E_n = (-2.18 \times 10^{-18})/n^2\ J$. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?

The electron energy in hydrogen atom is given by Eₙ = (-2.18 × 10⁻¹⁸)/n²J. Calculate the energy required to re

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E_n = (-2.18 \times 10^{-18})/n^2\ J

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Concept understanding — Energy Level Quantization

Energy Level Quantization: From Intuition to Precision

Imagine you're climbing a smooth ramp. You can stop at any height — 1 metre, 1.5 metres, 2.1 metres — anywhere you like. That's how we intuitively think about energy in everyday life: continuous, like a slide.

Now imagine a staircase. You can stand on step 1, step 2, or step 3 — but you cannot stand halfway between step 2 and step 3. There's no such place. The steps are discrete, not continuous.

Energy level quantization is the idea that in the microscopic world of atoms and molecules, energy behaves like a staircase, not a ramp. Electrons in an atom cannot have just any energy — they can only occupy specific, allowed energy levels. Everything else is forbidden.

Why does this happen? The core intuition

In the classical world, an electron orbiting a nucleus would continuously radiate energy, spiral inward, and crash — atoms would be unstable. But atoms are stable. Nature solved this problem by imposing a rule: the electron's angular momentum (and therefore its energy) can only take certain discrete values.

Think of a guitar string. It can only vibrate at specific frequencies — its fundamental and harmonics. You can't pluck it to produce a frequency halfway between two harmonics. The string's vibration is quantized by its boundaries. Similarly, an electron bound to a nucleus is confined in space, and that confinement forces its energy to be quantized.

Note

Quantization is not a mysterious extra rule — it emerges naturally whenever a wave (like an electron's matter wave) is confined. Confinement creates standing waves, and standing waves only exist at specific frequencies.

The precise statement

For a bound system (like an electron in an atom), the total energy $E$ of the system can only take certain discrete values:

E = E_1, E_2, E_3, \dots

where each $E_n$ is a specific, fixed number. The integer $n$ (1, 2, 3, …) is called the principal quantum number. The lowest energy level ( $n=1$ ) is the ground state; higher levels ( $n>1$ ) are excited states.

For the hydrogen atom, the allowed energies are given by:

E_n = -\frac{13.6 \text{ eV}}{n^2}

So:

$n=1$ : $E_1 = -13.6$ eV (ground state)
$n=2$ : $E_2 = -3.4$ eV
$n=3$ : $E_3 = -1.51$ eV
and so on, approaching 0 eV as $n \to \infty$ (the ionization limit)

Important

The negative sign means the electron is bound to the nucleus. Zero energy corresponds to the electron being free (ionized). The more negative the energy, the more tightly bound the electron.

How do we know this is real?

The most direct evidence comes from atomic spectra. When an electron jumps from a higher energy level to a lower one, it emits a photon of light with energy exactly equal to the difference:

\Delta E = E_{\text{higher}} - E_{\text{lower}} = hf

where $h$ is Planck's constant and $f$ is the frequency of the emitted light. …

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The energy required to remove an electron from the $n=2$ orbit of hydrogen is $5.45 \times 10^{-19}\ \text{J}$ , and the longest wavelength of light that can cause this transition is $3.65 \times 10^{-5}\ \text{cm}$ .

The key idea here is that "removing an electron completely" means taking it from a bound state ( $n=2$ ) to the ionization limit ( $n = \infty$ ), where the electron has zero total energy. The energy needed is simply the difference between the final energy ( $E_\infty = 0$ ) and the initial energy ( $E_2$ ). The longest wavelength corresponds to the smallest energy photon that can still cause this transition — that is, a photon with energy exactly equal to the ionization energy from $n=2$ .

Energy level quantization in hydrogen means the electron can only occupy specific orbits, each with a well-defined energy given by $E_n = -\frac{2.18 \times 10^{-18}}{n^2}\ \text{J}$ . The negative sign indicates that the electron is bound to the nucleus — you must add energy to free it. At $n = \infty$ , the electron is free and at rest, so $E_\infty = 0$ . The energy required to remove the electron from any level $n$ is therefore $0 - E_n = -E_n$ , which is positive.

Let's work through the calculation.

Find the energy of the electron in the $n=2$ orbit. Using the given formula:

E_2 = -\frac{2.18 \times 10^{-18}}{2^2} = -\frac{2.18 \times 10^{-18}}{4} = -5.45 \times 10^{-19}\ \text{J}

Calculate the ionization energy from $n=2$ . The energy required to remove the electron is the difference:

\Delta E = E_\infty - E_2 = 0 - (-5.45 \times 10^{-19}) = 5.45 \times 10^{-19}\ \text{J}

This is the minimum energy a photon must have to eject the electron from the $n=2$ level.

Relate photon energy to wavelength. The energy of a photon is given by $E = \frac{hc}{\lambda}$ , where $h = 6.626 \times 10^{-34}\ \text{J·s}$ is Planck's constant and $c = 3.00 \times 10^8\ \text{m/s}$ is the speed of light. Solving for wavelength:

\lambda = \frac{hc}{\Delta E}

Plug in the numbers.

\lambda = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{5.45 \times 10^{-19}}

First compute the numerator:

6.626 \times 10^{-34} \times 3.00 \times 10^8 = 1.9878 \times 10^{-25}\ \text{J·m}

Then divide:

$\lambda = \frac{1.9878 \times 10^{-25}}{5.45 \times 10^{-19}} = 3.647 \times 10^{-7}\ \text{m}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 16 on this concept.

TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
$\begin{array}{|c|c|} \hline \text{List-1 (Element)} & \text{List-2 (Block)} \\ \hline \text{A Hf} & \text{I s-block} \\ \text{B Ra} & \text{II p-Block} \\ \text{C Am} & \text{III d-Block} \\ \text{D At} & \text{IV f-Block} \\ \hline \end{array}$
›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:

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).

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).

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).

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: $[Xe] 4f^{14} 5d^{10} 6p^5$ . …
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TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
$^{2+}$
›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 $n=3$ , $Z=3$ :

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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.

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.

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.

Place the element from the next period. …
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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 $\phi$ .

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.

Compare each given wavelength to $\lambda_0$ . …
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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.

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

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

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.

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:

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

Match with the options …
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TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
$h = 6.62 \times 10^{-34}$
›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.

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

$L$ 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.

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 $n=1$ .

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 $\pi \approx 3.14159$ :

$L = \frac{6.62 \times 10^{-34}}{2 \times 3.14159} \text{ Js}$ …
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TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
$\lambda$
›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 $R=\dfrac{16}{3\lambda}$ . …
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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.

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.

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.

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: …
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TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
$\mathrm{Li}^{2+}$
›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.

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 $x = 3 a_0$ .

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

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

Express the $\mathrm{He}^{+}$ radius in terms of $x$ . Since $x = 3 a_0$ , we have $a_0 = x/3$ . Substitute: $r_{\mathrm{He}^{+}, n=4} = 8 \cdot \frac{x}{3} = \frac{8}{3} x$ …
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TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
$_2$
›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.

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.

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.

XeF $_6$ …
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TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
$h = 6.625 \times 10^{-34}$
›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.

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.

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

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

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

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

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

Match with the options …
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TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
$\mathrm{Na}^+ > \mathrm{Al}^{3+} > \mathrm{Mg}^{2+} > \mathrm{K}^+$
›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.

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.

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.

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.

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).

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.

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. …
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