MCQ
Total number of $d$ electrons present an element with atomic no. $78$ is
- A$8$
- B$58$
- C$28$
- ✓$29$
✓
Answer
Correct option: D.
$29$
d
${}_{78}Pt = 3{d^{10}} + 4{d^{10}} + 5{d^9} = 29\,{e^{ - 1}}s$
${}_{78}Pt = 3{d^{10}} + 4{d^{10}} + 5{d^9} = 29\,{e^{ - 1}}s$
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