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Question

What will be the value of n-factor (nf) or valence factor of P in Ca3P2 and H3PO4 respectively for the following change:

Ca3P2H3PO4

A
10,5
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B
4,8
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C
16,8
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D
3,3
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Solution

The correct option is C 16,8
Oxidation state of P in Ca3P2:
(2×3)+2x=0x=3
Oxidation state of P in H3PO4
(3×1)+x+(4×2)=0x=+5

Ca33P2H3+5PO4
Difference of Oxidation number =5(3)=8
nf=Difference in oxidation number×Number of atoms
nf of Ca3P2=8×2=16
nf of H3PO4=8×1=8

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