Question

Consider ‘n’ is the number of lone pair of electrons present in the equatorial position of the most stable structure of \( \text{ClF}_3 \). The ions from the following with ‘n’ number of unpaired electrons are:
A. \( \text{V}^{3+} \)  
B. \( \text{Ti}^{3+} \) 
C. \( \text{Cu}^{2+} \) 
D. \( \text{Ni}^{2+} \) 
E. \( \text{Ti}^{2+} \)
Choose the correct answer from the options given below:

• A. A, D and E Only
• B. B and C Only
• C. B and D Only
• D. A and C Only

Hint:

To determine the number of unpaired electrons, look at the electronic configurations of the ions.

Answer 1

The Correct Option is A

The molecule ClF3 has a T-shaped structure. Chlorine has 7 valence electrons. In ClF3, there are 3 bond pairs with fluorine and 2 lone pairs.

The most stable structure of ClF3 has the two lone pairs in the equatorial positions of a trigonal bipyramidal arrangement. Therefore, the number of lone pairs in the equatorial positions is n = 2.

Now we need to find which of the given ions have 2 unpaired electrons, matching the value of n = 2:

A. V3+: Vanadium (V) has electronic configuration [Ar] 3d3 4s2.
So V3+ has electronic configuration [Ar] 3d2.
It has 2 unpaired electrons.

B. Ti3+: Titanium (Ti) has electronic configuration [Ar] 3d2 4s2.
So Ti3+ has electronic configuration [Ar] 3d1.
It has 1 unpaired electron.

C. Cu2+: Copper (Cu) has electronic configuration [Ar] 3d10 4s1.
So Cu2+ has electronic configuration [Ar] 3d9.
It has 1 unpaired electron.

D. Ni2+: Nickel (Ni) has electronic configuration [Ar] 3d8 4s2.
So Ni2+ has electronic configuration [Ar] 3d8.
It has 2 unpaired electrons.

E. Ti2+: Titanium (Ti) has electronic configuration [Ar] 3d2 4s2.
So Ti2+ has electronic configuration [Ar] 3d2.
It has 2 unpaired electrons.

Conclusion: The ions with exactly 2 unpaired electrons are V3+, Ni2+, and Ti2+.

Final Answer: The final answer is (1) A, D and E only.

Answer 2

Step 1: Determine the structure and lone pairs in ClF₃.
Chlorine (Cl) has 7 valence electrons, and each fluorine (F) atom contributes one electron for bonding. Hence, the total number of electrons around chlorine is:
\[ 7 + 3 = 10 \text{ electrons (or 5 pairs).} \]
Three of these pairs are bonding pairs (from Cl–F bonds) and two are lone pairs. According to the VSEPR theory, ClF₃ has the formula AX₃E₂ and exhibits a T-shaped molecular geometry with two lone pairs.

Step 2: Identify lone pairs in the equatorial position.
In the trigonal bipyramidal electron geometry of ClF₃, lone pairs occupy equatorial positions to minimize repulsion. Thus, both lone pairs lie in the equatorial plane.
Therefore, the number of lone pairs in the equatorial position is:
\[ n = 2 \]

Step 3: Find ions with ‘n’ (2) unpaired electrons.
We now look for ions having 2 unpaired electrons:

(A) V³⁺: [Ar] 3d² → 2 unpaired electrons ✅
(B) Ti³⁺: [Ar] 3d¹ → 1 unpaired electron ❌
(C) Cu²⁺: [Ar] 3d⁹ → 1 unpaired electron ❌
(D) Ni²⁺: [Ar] 3d⁸ → 2 unpaired electrons ✅
(E) Ti²⁺: [Ar] 3d² → 2 unpaired electrons ✅

Step 4: Conclusion.
The ions having 2 unpaired electrons (equal to n = 2) are:
V³⁺, Ni²⁺, and Ti²⁺.

Final Answer:
A, D and E only