Question

If the no. of orbitals of a particular type were (3l + 1), but spin quantum no.s only +1/2 and -1/2, then a d-type orbital will contain how many electrons?

Answer 1

For d-orbital, l = 2

Then according to the eqn- (2l + 1)

No of d - orbitals = (2*2+ 1)

= 4 + 1

= 5 orbitals.

Each orbital contain 2 electrons, one having +1/2 spin and the other with -1/2 spin.

Hence a d - type orbital contains 2 x 5 = 10 electrons.

Answer 2

Answer:

$l = 2 \: \: for \: d \: orbital$

$so \: given \: no.of \: orbitals = (3l + 1)$

$since \: l = 2$

$no.of \: orbitals = (3 \times 2 + 1)$

$= 7$

$no.of \: possible \: spin \: quantum \: no.s = 2$

$so \: no.of \: electrons = 2 \times 7 = 14electrons$