The compound of vanadium has magnetic moment of 1.73 BM. The vanadium chloride has the formula

Question

The compound of vanadium has magnetic moment of 1.73 BM. The vanadium chloride has the formula (A) VCl 2 (B) VCl 3 (C) VCl 4 (D) VCl 5

Tardigrade Admin · Accepted Answer

V (Z=23)=[A r] 4 s2 3 d3 V x +=? <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/49fa33764b1453d65fcdd7f08d1fab7c-.png" /> If there are N unpaired electrons then Magnetic moment =√N(N+2) B M=1.73 ∴ N(N+2)=3 ∴ N=1 Thus, vanadium exists as V4+ V4+=[A r] 3 d1 Thus, vanadium chloride is VCl 4

Q. The compound of vanadium has magnetic moment of $1.73$ BM. The vanadium chloride has the formula

$V C l_{2}$

$V C l_{3}$

$V C l_{4}$

$V C l_{5}$

$V (Z = 23) = [A r] 4 s^{2} 3 d^{3}$
$V^{x +} = ?$

If there are $N$ unpaired electrons then
Magnetic moment $= N (N + 2) BM = 1.73$
$∴ N (N + 2) = 3$
$∴ N = 1$
Thus, vanadium exists as $V^{4 +}$
$V^{4 +} = [A r] 3 d^{1}$
Thus, vanadium chloride is $V C l_{4}$

Q. The compound of vanadium has magnetic moment of 1.73 BM. The vanadium chloride has the formula

VCl2​

VCl3​

VCl4​

VCl5​