wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
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
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
4,8
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
16,8
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
3,3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
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

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon