Question

question 3 of 32 what is the electron configuration for vanadium (v)? the periodic table a. 1s²2s²2p⁶3s²3p⁶4s²4d³ b. 1s²2s²2p⁶3s²3p⁶4s²3d³ c. 1s²2s²2p⁶3s²3p⁶3d⁵ d. 1s²2s²2p⁶3s²3p⁶4s²4p³

Explanation:

Step1: Determine vanadium's atomic number

Vanadium has an atomic number of 23, so it has 23 electrons.

Step2: Apply electron - filling rules

Electrons fill orbitals in the order 1s, 2s, 2p, 3s, 3p, 4s, 3d. The s - subshell can hold 2 electrons, the p - subshell can hold 6 electrons, and the d - subshell can hold 10 electrons.
1s gets 2 electrons: $1s^{2}$
2s gets 2 electrons: $2s^{2}$
2p gets 6 electrons: $2p^{6}$
3s gets 2 electrons: $3s^{2}$
3p gets 6 electrons: $3p^{6}$
4s gets 2 electrons before 3d (due to lower energy initially): $4s^{2}$
The remaining 3 electrons go into 3d: $3d^{3}$
The electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{3}$

Answer:

B. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{3}$