Q.Calculate the number of protons, neutrons and electrons in $^{80}_{35}Br$.

The key here is understanding how the periodic table is built. Atomic number 114 is far beyond the elements you normally study, but the periodic table's structure repeats in blocks — s, p, d, f — so you can figure out its group just by counting.

Every element in a given group has the same outer shell electron configuration (the same number of valence electrons in the same type of orbital). So if element 114 has the same outer configuration as element X, they must belong to the same group. The question is: which group?

Let’s work it out.

1. Find the period of element 114.

   The periods fill in order:

   • Period 1: 2 elements (H, He)
   • Period 2: 8 elements (Li–Ne)
   • Period 3: 8 elements (Na–Ar)
   • Period 4: 18 elements (K–Kr)
   • Period 5: 18 elements (Rb–Xe)
   • Period 6: 32 elements (Cs–Rn) — includes the 4f block (14 lanthanides)
   • Period 7: 32 elements (Fr–Og) — includes the 5f block (14 actinides)

   Cumulative count up to period 6:

   $2 + 8 + 8 + 18 + 18 + 32 = 86$ (that’s radon, Rn).

   So element 114 is in period 7, because 86 + 32 = 118, and 114 falls within that range.

2. Locate element 114 within period 7.

   Period 7 starts with Fr (87) and Ra (88). Then come the 5f elements (actinides): atomic numbers 89–102 (14 elements). After that, the 6d block starts.

   • 87–88: s-block (Groups 1 and 2)
   • 89–102: f-block (actinides) — these are placed below the main table
   • 103–112: d-block (transition metals, Groups 3–12)
   • 113–118: p-block (Groups 13–18)

   So element 114 is in the p-block. Count from 113 (Group 13):

   • 113: Group 13
   • 114: Group 14
   • 115: Group 15
   • 116: Group 16
   • 117: Group 17
   • 118: Group 18

   Therefore, element 114 is in Group 14 (the carbon family).

3. Identify element X with the same outer configuration.

   Group 14 elements have the outer electron configuration $ns^2 np^2$. The common members are C, Si, Ge, Sn, Pb. Among the options:

   • Ar (Group 18) — $3s^2 3p^6$
   • Ge (Group 14) — $4s^2 4p^2$
   • Se (Group 16) — $4s^2 4p^4$
   • Ti (Group 4) — $4s^2 3d^2$

   Only Ge is in Group 14. So element 114 has the same outer shell configuration as germanium.

The key idea is that “empirical formula weight” means the mass of the simplest ratio of elements, not the molecular formula. Borazine’s molecular formula is B₃N₃H₆ (a cyclic compound analogous to benzene), but the empirical formula is the reduced ratio: B : N : H = 1 : 1 : 2, i.e., BNH₂.

Why this works:

The empirical formula is the smallest whole‑number ratio of atoms in a compound. The empirical formula weight is just the sum of the atomic masses for that ratio. Many students mistakenly use the molecular formula (B₃N₃H₆) and get a larger number, but the question specifically asks for the empirical formula weight.

Let’s work it out:

1. Identify the molecular formula of borazine.

   Borazine is B₃N₃H₆ — three boron, three nitrogen, and six hydrogen atoms.

2. Find the simplest whole‑number ratio.

   Divide all subscripts by the greatest common factor, which is 3:

So the empirical formula is $\text{BNH}_2$.

3. Calculate the empirical formula weight.

   Use the given atomic masses:

   • Boron: $10.8 \, \text{u}$
   • Nitrogen: $14 \, \text{u}$
   • Hydrogen: $1 \, \text{u} \times 2 = 2 \, \text{u}$

   Sum:

4. Match with the options. The result 26.8 u corresponds to option (C).

The key idea is the Bohr radius formula for hydrogen-like atoms:

where $a_0$ is the Bohr radius (the $n=1$ radius for hydrogen, $Z=1$).

For hydrogen, $Z=1$; for He⁺, $Z=2$.

We are told the $n=2$ hydrogen radius is $x$ pm. That gives us a direct relation to $a_0$. Then we compute the $n=3$ He⁺ radius in terms of $x$.

1. Write the general formula For a hydrogen-like atom with nuclear charge $Z$, the radius of the $n$-th stationary state is

where $a_0 \approx 52.9$ pm is the Bohr radius (the $n=1$ radius for H).

2. Use the given data for hydrogen For hydrogen ($Z=1$), the $n=2$ radius is

This is given as $x$ pm, so

3. Find the radius for He⁺ For He⁺ ($Z=2$), the $n=3$ radius is

Substitute $a_0 = x/4$:

4. Check the options The result $\frac{9x}{8}$ matches option (B).

The question tests a straightforward factual recall from environmental chemistry: the relative abundance of noble gases in the atmosphere. Noble gases are all present in trace amounts, but their proportions vary significantly. Argon dominates because it is produced by the radioactive decay of potassium-40 in the Earth's crust and is chemically inert, so it accumulates over time.

1. Recall the atmospheric composition of noble gases.

   Dry air (by volume) contains roughly:

   • Argon: ~0.934%
   • Neon: ~0.0018%
   • Helium: ~0.00052%
   • Krypton: ~0.00011%
   • Xenon: ~0.000009%

   Argon is clearly the most abundant, by a factor of about 500 over the next most common (neon).

2. Eliminate the other options.

   • Helium (A) is abundant in the universe but scarce in the atmosphere because it escapes Earth's gravity.
   • Neon (B) is less than 0.002% — far behind argon.
   • Krypton (D) is even rarer, at about 0.0001%.

3. Confirm the reasoning.

   Argon's high abundance is due to its continuous production from the decay of $^{40}\text{K}$ (half-life ~1.25 billion years) and its inertness, which prevents it from being removed by chemical reactions. No other noble gas has such a significant terrestrial source.

The octet rule is a fundamental guideline in chemical bonding: atoms tend to gain, lose, or share electrons to achieve a stable configuration of 8 valence electrons (like a noble gas). Hydrogen is an exception — it needs only 2 electrons (a duplet). When checking a molecule, you must count the total electrons around each atom, including both bonding pairs and lone pairs. A molecule "obeys the octet rule" only if every atom (except H) has exactly 8 electrons in its valence shell.

Let’s examine each option one by one.

1. Option (A): $\mathrm{BCl}_3$

   Boron is in group 13 and has 3 valence electrons. It forms three single bonds with three chlorine atoms. Each bond contributes 1 electron from boron and 1 from chlorine, so boron gets 3 (its own) + 3 (from bonds) = 6 electrons. That’s an incomplete octet — boron is electron-deficient. Chlorine atoms each have 7 valence electrons; after one bond, they have 6 lone pairs (2 electrons each) plus the bonding pair, totalling 8. So chlorine obeys the octet, but boron does not. Hence, $\mathrm{BCl}_3$ fails.

2. Option (B): $\mathrm{BeH}_2$

   Beryllium has 2 valence electrons and forms two single bonds with hydrogen. Each bond gives beryllium 1 electron from each hydrogen, so beryllium ends up with 2 (its own) + 2 (from bonds) = 4 electrons. That’s far short of 8 — beryllium is another common exception (it often forms stable compounds with only 4 electrons). Hydrogen, with one bond, has 2 electrons (duplet satisfied). So $\mathrm{BeH}_2$ does not obey the octet for Be.

3. Option (C): $\mathrm{SF}_6$

   Sulfur has 6 valence electrons. It forms six single bonds with six fluorine atoms. Each bond gives sulfur 1 electron from each fluorine, so sulfur gets 6 (its own) + 6 (from bonds) = 12 electrons. This is an expanded octet — sulfur can use its d-orbitals to accommodate more than 8 electrons. Fluorine atoms each have 7 valence electrons; after one bond, they have 6 lone pairs (8 electrons total). So fluorine obeys the octet, but sulfur does not. $\mathrm{SF}_6$ violates the octet rule.

4. Option (D): $\mathrm{SCl}_2$

   Sulfur has 6 valence electrons. It forms two single bonds with two chlorine atoms. Each bond gives sulfur 1 electron from each chlorine, so sulfur gets 6 (its own) + 2 (from bonds) = 8 electrons. That’s a perfect octet. Additionally, sulfur has two lone pairs (4 electrons) left over, which are already counted in its 6 original electrons — the total around sulfur is 8. Each chlorine has 7 valence electrons; after one bond, it has 6 lone pairs (3 lone pairs × 2 electrons = 6) plus the bonding pair, totalling 8. All atoms — sulfur and both chlorines — have complete octets. $\mathrm{SCl}_2$ obeys the octet rule.

The concept first

Isoelectronic species have the same number of electrons but different nuclear charges. Since the electron count — and hence the screening — is identical, the only variable left is $Z$. The size of the ion is then governed by

with $\sigma$ (screening) essentially the same across the series. A larger $Z$ therefore means a larger pull on the same electron cloud, and a smaller ion.

Step 1 — Check the electron counts

Both have the configuration $1s^{2}2s^{2}2p^{6}$ (neon-like). The Reason (R) is correct.

Step 2 — Compare the nuclear charges

Ten electrons held by $+11$ are squeezed harder than ten electrons held by $+9$.

Step 3 — Compare the radii

The two radii are far from equal, so the Assertion (A) is incorrect. The general trend across an isoelectronic series (all $10$ electrons) is

radius falling steadily as $Z$ increases.

Step 4 — Map to the options

(A) false, (R) true — that is option (D).

ANSWER: D

Concept & Intuition

The Bohr model gives the radius of an electron orbit in a hydrogen-like atom as proportional to $n^2$ (the square of the principal quantum number) and inversely proportional to the nuclear charge $Z$. For hydrogen ($Z=1$), the first orbit ($n=1$) has a specific radius called the Bohr radius $a_0$. The problem asks: for which hydrogen-like ion does some orbit $n$ have exactly this same radius? That means the factor $\frac{n^2}{Z}$ must equal $1$. So we need $n^2 = Z$. Among the options, we look for a pair $(n, Z)$ where $Z$ is the atomic number of the ion (after removing all electrons but one) and $n$ is a small integer.

Step-by-step reasoning

1. Recall the Bohr radius formula For a hydrogen-like atom (one electron, nuclear charge $Ze$), the radius of the $n$th orbit is

where $a_0 = 0.529 \, \text{Å}$ is the Bohr radius (the radius of the $n=1$ orbit in hydrogen).

2. Set the condition We want $r_n$ (for the ion X) to equal $a_0$ (the first Bohr orbit of H). So:

3. Interpret $Z$ for the ion

   The ion X is hydrogen-like, meaning it has only one electron. Its nuclear charge $Z$ is the atomic number of the element. For example:

   • He⁺: $Z = 2$
   • Li²⁺: $Z = 3$
   • Be³⁺: $Z = 4$

4. Check each option

   • (A) $n=2$, Li²⁺ ($Z=3$): $n^2 = 4$, $Z=3$ → not equal.
   • (B) $n=3$, Li²⁺ ($Z=3$): $n^2 = 9$, $Z=3$ → not equal.
   • (C) $n=2$, Be³⁺ ($Z=4$): $n^2 = 4$, $Z=4$ → equal!
   • (D) $n=2$, He⁺ ($Z=2$): $n^2 = 4$, $Z=2$ → not equal.

5. Conclusion

   Only option (C) satisfies $n^2 = Z$. So the orbit number is $n=2$ and the species is Be³⁺.

The key concept here is the systematic naming of superheavy elements by IUPAC. Elements with atomic numbers 103 and above have official one- or three-letter symbols that are not always obvious from their English names. The trick is to recall the specific symbols for elements 103 (Lawrencium), 107 (Bohrium), 109 (Meitnerium), and 111 (Roentgenium). A common pitfall is confusing the symbols for elements 107–111, which all end in “-ium” and have similar-looking abbreviations.

Let’s match each atomic number to its correct IUPAC symbol step by step.

1. Atomic number 103 – This is Lawrencium, named after Ernest Lawrence. Its IUPAC symbol is Lr. In the list, that corresponds to III.

   So A → III.

2. Atomic number 107 – This is Bohrium, named after Niels Bohr. Its symbol is Bh. That matches II.

   So B → II.

3. Atomic number 109 – This is Meitnerium, named after Lise Meitner. Its symbol is Mt. That matches I.

   So C → I.

4. Atomic number 111 – This is Roentgenium, named after Wilhelm Röntgen. Its symbol is Rg. That matches V.

   So D → V.

Now assemble the full mapping:

A–III, B–II, C–I, D–V.

Looking at the options:

• (A) says A–II, B–III, C–V, D–IV → wrong.
• (B) says A–III, B–I, C–II, D–IV → wrong.
• (C) says A–II, B–V, C–IV, D–I → wrong.
• (D) says A–III, B–II, C–I, D–V → correct.

Concept and Intuition

Hydrogen has three naturally occurring isotopes: protium (¹H), deuterium (²H or D), and tritium (³H). Their differences come from the number of neutrons in the nucleus. Protium has just one proton and zero neutrons; deuterium has one proton and one neutron; tritium has one proton and two neutrons. The question tests basic knowledge of these differences, especially relative abundance, atomic mass, neutron count, and radioactivity. The most straightforward fact is that deuterium has one more neutron than protium.

Step-by-step reasoning

1. Relative abundance – Protium makes up over 99.98% of all hydrogen atoms on Earth; deuterium is only about 0.0156%. So deuterium is far less abundant than protium. Option (A) is false.

2. Relative atomic mass – The atomic mass of protium is approximately 1.0078 u, while deuterium is about 2.0141 u. Deuterium is indeed heavier, so its relative atomic mass is greater. However, the phrasing “relative atomic mass” here compares the isotopes themselves, not their average in nature. This statement is true in isolation, but we must check if it is the correct statement among the options — and we will see that another option is more directly and unambiguously correct.

3. Neutron count – Protium has 0 neutrons; deuterium has 1 neutron. Therefore, deuterium has more neutrons than protium. This is a simple, undeniable fact. Option (C) is true.

4. Radioactivity – Deuterium is stable; it does not undergo radioactive decay. Tritium is radioactive (half-life ~12.3 years), but deuterium is not. Option (D) is false.

Conclusion – The only statement that is both true and unambiguous is that deuterium has more neutrons than protium.