Arrange the following species in the increasing order of their magnetic moment.
$(I) V C l_{3} (I I) V O S O_{4} (I I I) N a_{3} V O_{4} (I V) [V (H_{2} O)_{6}] S O_{4} . H_{2} O$

I I I < I I < I < I V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

I I I < I V < I I < I

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I I < I I I < I < I V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I V < I < I I < I I I

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $I I I < I I < I < I V$
The compound which has the maximum number of unpaired electrons will have higher magnetic moment.
The complexes and the number of unpaired electrons present in them are given below:
$V C l_{3}; V^{3 +}; [A r]^{18} 3 d^{2} = 2$ unpaired electrons
$V O^{2 +}; V^{4 +}; [A r]^{18} 3 d^{1} = 1$ unpaired electron
$[V O_{4}]^{3 -}; V^{5 +}; [A r]^{18} 3 d^{0} =$ no unpaired electron
$[V (H_{2} O)_{6}]^{2 +}; V^{2 +}; [A r]^{18} 3 d^{3} = 3$ unpaired electrons
Hence, the correct order is $I I I < I I < I < I V$ .

Suggest Corrections

Similar questions

V C l_{3}

V O S O_{4}

N a_{3} V O_{4}

$[V {(H_{2} O)}_{6}] S O_{4} . H_{2} O$

C e^{3 +}

L a^{3 +}

P m^{3 +}

Y b^{3 +}

Arrange the following in the increasing order of their basicity.

CH₃NH₂(I) ;(CH₃)₂NH(II) ;C₆H₅NH₂(III) ;(CH₃)₃N(IV)

C H_{3} C H_{2} C H_{2} B r

C H_{3} C H_{2} C H_{2} C H_{2} B r

(C H_{3})_{2} C H C H_{2} B r

(C H_{3})_{3} C B r

Join BYJU'S Learning Program