Among $[N i (C O)_{4}]$ , $[N i C l_{4}]^{2 -}$ , $[C o (N H_{3})_{4} C l_{2}] C l$ , $N a_{3} [C o F_{6}]$ , $N a_{2} O_{2}$ and $C s O_{2}$ , the total number of paramagnetic compounds are:

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Solution

The correct option is B 3

Compound/Ion Magnetic nature of the
compound

$[N i C O_{4}]$ Diamagnetic

$[N i C l_{4}]^{2 -}$ Paramagnetic

$[C o (N H_{3})_{4} C l_{2}] C l$ Diamagnetic

$N a_{3} [C o F_{6}]$ Paramagnetic

$N a_{2} O_{2}$ Diamagnetic

$C s O_{2}$ Paramagnetic

The number of paramagnetic compounds is 3.

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Similar questions

Among $[N i (C o)_{4}], [N i C l_{4}]^{2 -}, [C o (N H_{3})_{4} C l_{2}] C l, N a_{2} [C o F_{6}]^{3 +}, N a_{2} O_{2} a n d C s O_{2},$ the total number of paramagnetic compounds is

Q. Amongst

[N i C l_{4}]^{2 -}, [N i (H_{2} O)_{6}]^{2 +}, [N i (P P h_{3})_{2} C l_{2}], [N i (C O)_{4}]

and

[N i (C N)_{4}]^{2 -}

, the paramagnetic species are :

[NiCl₄]²⁻ is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why?

[N i C l_{4}]

is paramagnetic while

[N i C O_{4}]

is diamagnetic though both are tetrahedral. Explain.

Q. Among

[N i (C O)_{4}], [C i (C N)_{4}]^{2 -}

and

[N i (C l)_{4}]^{2 -}

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Compound/Ion	Magnetic nature of the compound
$[N i C O_{4}]$	Diamagnetic
$[N i C l_{4}]^{2 -}$	Paramagnetic
$[C o (N H_{3})_{4} C l_{2}] C l$	Diamagnetic
$N a_{3} [C o F_{6}]$	Paramagnetic
$N a_{2} O_{2}$	Diamagnetic
$C s O_{2}$	Paramagnetic

Among [Ni(CO)4], [NiCl4]2−, [Co(NH3)4Cl2]Cl, Na3[CoF6], Na2O2 and CsO2, the total number of paramagnetic compounds are:

The correct option is B 3 Compound/Ion Magnetic nature of the compound [NiCO4] Diamagnetic [NiCl4]2− Paramagnetic [Co(NH3)4Cl2]Cl Diamagnetic Na3[CoF6] Paramagnetic Na2O2 Diamagnetic CsO2 Paramagnetic The number of paramagnetic compounds is 3.

Among $[N i (C O)_{4}]$ , $[N i C l_{4}]^{2 -}$ , $[C o (N H_{3})_{4} C l_{2}] C l$ , $N a_{3} [C o F_{6}]$ , $N a_{2} O_{2}$ and $C s O_{2}$ , the total number of paramagnetic compounds are: