Chloro compound of Vanadium has only spin magnetic moment of $$1.73\,BM.$$ This Vanadium chloride has the formula :
A.
$$VC{l_2}$$
B.
$$VC{l_4}$$
C.
$$VC{l_3}$$
D.
$$VC{l_5}$$
Answer :
$$VC{l_4}$$
Solution :
$$\eqalign{
& \mu = \sqrt {n\left( {n + 2} \right)} \cr
& 1.73 = \sqrt {n\left( {n + 2} \right)} \cr} $$
On calculating the value of $$n$$ we find $$n = 1$$
No. of unpaired electrons $$= 1$$
hence its configuration will be
$$\eqalign{
& V\left( {23} \right) = \left[ {Ar} \right]3{d^3}4{s^2} \cr
& {V^{4 + }} = \left[ {Ar} \right]3{d^1} \cr} $$
∴ Its chloride has the formula $$VC{l_4}$$
Releted MCQ Question on Inorganic Chemistry >> D and F Block Elements
Releted Question 1
When same amount of zinc is treated separately with excess of sulphuric acid and excess of sodium hydroxide, the ratio of volume of hydrogen evolved is