Question

1. write the electron configuration for the following elements.

a) s
b) li
c) ca
d) zn

1. write the noble gas configuration for the following elements.

a) sc
b) br
c) pb
d) mg

1. identify the following elements according to their electron configuration.

a) ( 1 s ^ { 2 } 2 s ^ { 2 } 2 p ^ { 6 } 3 s ^ { 2 } 3 p ^ { 6 } 4 s ^ { 2 } 3 d ^ { 3 } )
b) ( 1 s ^ { 2 } 2 s ^ { 2 } 2 p ^ { 6 } 3 s ^ { 2 } 3 p ^ { 1 } )
c) ( 1 s ^ { 2 } 2 s ^ { 2 } 2 p ^ { 6 } 3 s ^ { 2 } 3 p ^ { 6 } 4 s ^ { 2 } 3 d ^ { 10 } 4 p ^ { 6 } 5 s ^ { 2 } 4 d ^ { 10 } 5 p ^ { 6 } 6 s ^ { 2 } 4 f ^ { 14 } 5 d ^ { 10 } 6 p ^ { 6 } 7 s ^ { 2 } 5 f ^ { 14 } 6 d ^ { 5 } )
d) ( 1 s ^ { 2 } 2 s ^ { 2 } 2 p ^ { 6 } 3 s ^ { 2 } 3 p ^ { 6 } 4 s ^ { 2 } 3 d ^ { 10 } 4 p ^ { 2 } )

Explanation:

Step1: Determine the atomic number of each element

The atomic number of an element is equal to the number of electrons in a neutral atom.

• For \(S\) (sulfur), the atomic number \(Z = 16\).
• For \(Li\) (lithium), \(Z=3\).
• For \(Ca\) (calcium), \(Z = 20\).
• For \(Zn\) (zinc), \(Z=30\).
• For \(Sc\) (scandium), \(Z = 21\).
• For \(Br\) (bromine), \(Z=35\).
• For \(Pb\) (lead), \(Z = 82\).
• For \(Mg\) (magnesium), \(Z=12\).

Step2: Write the electron configuration using the Aufbau principle

The Aufbau principle states that electrons fill orbitals in order of increasing energy: \(1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d\cdots\)

• For \(S\) (\(Z = 16\)):

Electrons fill as follows: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}\)

• For \(Li\) (\(Z = 3\)):

\(1s^{2}2s^{1}\)

• For \(Ca\) (\(Z = 20\)):

\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}\)

• For \(Zn\) (\(Z = 30\)):

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

Step3: Write the noble - gas configuration

Noble - gas configuration is a shorthand method. We replace the electron configuration of the previous noble gas with the noble gas symbol.

• For \(Sc\) (\(Z = 21\)):

The previous noble gas is \(Ar\) (\(Z = 18\)). So, \([Ar]4s^{2}3d^{1}\)

• For \(Br\) (\(Z = 35\)):

The previous noble gas is \(Ar\). So, \([Ar]4s^{2}3d^{10}4p^{5}\)

• For \(Pb\) (\(Z = 82\)):

The previous noble gas is \(Xe\) (\(Z = 54\)). So, \([Xe]6s^{2}4f^{14}5d^{10}6p^{2}\)

• For \(Mg\) (\(Z = 12\)):

The previous noble gas is \(Ne\) (\(Z = 10\)). So, \([Ne]3s^{2}\)

Step4: Identify the element from the electron configuration

Count the total number of electrons (sum of the exponents in the electron - configuration). The number of electrons in a neutral atom is equal to the atomic number.

• For \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{3}\):

Sum of exponents \(=2 + 2+6 + 2+6 + 2+3=23\). The element with \(Z = 23\) is \(V\) (vanadium).

• For \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}\):

Sum of exponents \(=2+2 + 6+2+1=13\). The element with \(Z = 13\) is \(Al\) (aluminum).

• For \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^{10}6p^{6}7s^{2}5f^{14}6d^{5}\):

Sum of exponents \(=2+2+6+2+6+2+10+6+2+10+6+2+14+10+6+2+14+5 = 107\). The element with \(Z = 107\) is \(Bh\) (bohrium).

• For \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{2}\):

Sum of exponents \(=2+2+6+2+6+2+10+2=32\). The element with \(Z = 32\) is \(Ge\) (germanium).

Answer:

3)
a) \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}\)
b) \(1s^{2}2s^{1}\)
c) \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}\)
d) \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}\)

4)
a) \([Ar]4s^{2}3d^{1}\)
b) \([Ar]4s^{2}3d^{10}4p^{5}\)
c) \([Xe]6s^{2}4f^{14}5d^{10}6p^{2}\)
d) \([Ne]3s^{2}\)

5)
a) \(V\)
b) \(Al\)
c) \(Bh\)
d) \(Ge\)