If Pauli exclusion principle allows three electron an orbital, and other rules are remaining same then find the block of Co(.Z=27.)

Question

If Pauli exclusion principle allows three electron an orbital, and other rules are remaining same then find the block of Co(.Z=27.) (A) d (B) p (C) s (D) can't be predicted

Tardigrade Admin · Accepted Answer

Co(.27.):-1s3,2s3,2p9,3s3,3p9 (If one orbital allows three electrons)

Q. If Pauli exclusion principle allows three electron an orbital, and other rules are remaining same then find the block of $C o (Z = 27)$

$d$

$p$

$s$

can't be predicted

$C o (27) : - 1 s^{3}, 2 s^{3}, 2 p^{9}, 3 s^{3}, 3 p^{9}$ (If one orbital allows three electrons)

Q. If Pauli exclusion principle allows three electron an orbital, and other rules are remaining same then find the block of Co(Z=27)

d

p

s

can't be predicted

Co(27):−1s3,2s3,2p9,3s3,3p9 (If one orbital allows three electrons)

Q. If Pauli exclusion principle allows three electron an orbital, and other rules are remaining same then find the block of $C o (Z = 27)$

$d$

$p$

$s$

$C o (27) : - 1 s^{3}, 2 s^{3}, 2 p^{9}, 3 s^{3}, 3 p^{9}$ (If one orbital allows three electrons)