A neutral copper atom undergoes oxidation to form a copper(II) ion, Cu^{2+}. Given that th… — JAMB UTME

A neutral copper atom undergoes oxidation to form a copper(II) ion, $Cu^{2+}$ . Given that the atomic number of copper ( $Cu$ ) is 29, which of the following represents the correct ground-state electronic configuration of $Cu^{2+}$ ?

$1s^2 2s^2 2p^6 3s^2 3p^6 3d^9$ Answer
B
$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7$
C
$1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^8$
D
$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^9$

Answer

The correct ground-state electronic configuration of the copper(II) ion (

Cu^{2+}

) is

1s^2 2s^2 2p^6 3s^2 3p^6 3d^9

The correct option correctly applies the rules of electronic configuration for transition metals. Neutral copper (

Z=29

) has an ground state of

1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1

. When ionized to

Cu^{2+}

, two electrons are removed: the outermost

4s^1

electron is lost first, followed by one electron from the

3d^{10}

subshell, leaving

1s^2 2s^2 2p^6 3s^2 3p^6 3d^9

Step-by-Step Solution

Determine the total number of electrons in neutral copper (

Cu

) and write its ground-state configuration.

Atomic number

Z = 29

, so neutral

Cu

has 29 electrons. Due to the extra stability of a completely filled

d

-subshell, the configuration is

1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1

Full subshell stability causes one electron from the

4s

orbital to promote to the

3d

subshell in neutral copper.

Calculate the number of electrons lost to form the

Cu^{2+}

ion.

Cu^{2+}

loses 2 electrons, leaving

29 - 2 = 27

electrons.

+2

charge indicates the loss of two valence electrons.

Remove two electrons following the rule for transition metal cation formation.

Remove the 1 electron from the outermost

4s

orbital first, then 1 electron from the

3d

orbital, leaving

3d^9

Electrons in the highest principal energy level (

n=4

) are lost before inner

(n-1)d

electrons (

n=3

Key Concept

Electronic configuration of d-block cations

Estimated Time:1m 0s

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