Question

magnetic moment of an ion m3+ is ✓35 bm. then no. of electron in d orbital of an element will be
(1.) 2
(2.) 5
(3.) 7
(4.) 6

Answer 1

Answer:

option 4

Explanation:

$\sqrt{5 \times 7}$

so unpaired electrons are 5 after removing 3 electrons 2 from 4s 1 from d

so total 6 electrons

Answer 2

The no. of electron in d orbital of an element will be 6

Explanation:

Formula used for magnetic moment :

$\mu=\sqrt{n(n+2)}$

where,

$\mu$ = magnetic moment

n = number of unpaired electrons

$\sqrt{35}=\sqrt{n(n+2)}$

${35}={n(n+2)}$

$n=5$

Thus then no. of unpaired electrons of an element will be 5.

As the metal is $M^{3+}$, the number of electrons removed are 3 , i.e 2 from s subshell and 1 from d subshell. Thus the number of electrons in d orbital must be 6 in metal M

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