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. \( \text{V}^{3+} \)
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:
Show Hint
- A, D and E Only
- B and C Only
- B and D Only
- A and C Only
The Correct Option is A
Approach Solution - 1
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.
Approach Solution -2
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
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