Question

Problem 3. (8 points) Write out the electron configuration for each of the following metal ions. (а) К* (b) Cа* (c) Sc3+ F7 (d) Ti* Problem 4. (8 points) Write out the electron configuration for each of the following nonmetal ions (a)CI (b)s2- (c) p3- (d)Sit Problem 5 (6 noints)

Answer 1

3.

(a) $K^+$

K has an atomic number of 19.

Hence, neutral K has 19 electrons. Now using aufbau principle, we can write the electronic configuration of neutral K atom as follows:

K:1s 2s 2p 3s 3p 4s

To form the K⁺ ion, the neutral K atom needs to lose 1 electron. The electron is first lost from the outermost shell. Hence, the 4s electron will be lost to form K⁺ ion.

Hence the electronic configuration of K⁺ ion can be written as

K+: 1s 2s 2p 3s 3p

(b) $Ca^{2+}$

Ca has atomic number 20.

Hence, neutral Ca has 20 electrons. Now using aufbau principle, we can write the electronic configuration of neutral Ca atom as follows:

Ca : 1s2 2s2 2p6 3s2 3pნ 45-

To form the Ca²⁺ ion, the neutral Ca atom needs to lose 2 electron. The electron is first lost from the outermost shell. Hence, the 4s electrons will be lost to form Ca²⁺ ion.

Hence the electronic configuration of Ca²⁺ ion can be written as

Ca2+ : 1s2 2s2 2p6 3s2 3pნ

(c) $Sc^{3+}$

Sc has atomic number 21.

Hence, neutral Sc has 21 electrons. Now using aufbau principle, we can write the electronic configuration of neutral Sc atom as follows:

Sc: 1s2 2s2 2p6 3s2 3pნ 4s2 3d1

To form the Sc³⁺ ion, the neutral Sc atom needs to lose 3 electron. First two electrons will be removed from outermost 4s orbital. The third electron will be lost from the next outermost orbital 3d.

Hence the electronic configuration of $Sc^{3+}$ ion can be written as

s3+ : 12,2 2p6 3s2 3pნ

(d) $Ti^{4+}$

Ti has atomic number 22.

Hence, neutral Ti has 22 electrons. Now using aufbau principle, we can write the electronic configuration of neutral Ti atom as follows:

Ti : 1s2 2s2 2p6 3s2 3pნ 452 3d2

To form the Ti⁴⁺ ion, the neutral Ti atom needs to lose 4 electron. First two electrons will be removed from outermost 4s orbital. The next two electrons will be lost from the next outermost orbital 3d.

Hence the electronic configuration of $Ti^{4+}$   ion can be written as

Ti4+ : 1s2 2s2 2p6 3s2 3p-

Note that each of the above ions have same electronic configuration and they are known as isoelectronic species.

4.

(a) $Cl^-$

Cl has atomic number 17.

Hence, neutral Cl atom has 17 electrons. Hence, the electronic configuration of neutral Cl atom is

Ci : 1s2 2s2 2p6 3s2 3p-

Now, to form Cl^- ion, we need to add one electron to the nearest vacant orbital of the neutral Cl atom. Now, the first vacancy is in the 3p orbital. Hence, the incoming electron will be added to 3p orbital. Note that 3p orbital can hold a total of 6 electrons.

Hence, the configuration of Cl^- ion is

Cl-:18 2s 2p 3s 3p

(b) $S^{2-}$

S has atomic number 16.

Hence, neutral Cl atom has 16 electrons. Hence, the electronic configuration of neutral S atom is

S: 152 2s 2p 3s 3p+

Now, to form S^2- ion, we need to add two electron to the nearest vacant orbital of the neutral Cl atom. Now, the first vacancy is in the 3p orbital. Hence, the incoming electron will be added to 3p orbital.

Hence, the configuration of S^2- ion is

s2– : 1s2 2s2 2p6 3s2 3p-

(c) $P^{3-}$

P has atomic number 15.

Hence, neutral P atom has 15 electrons. Hence, the electronic configuration of neutral P atom is

P: 1s2 2s2 2p6 3s2 3p-

Now, to form $P^{3-}$ ion, we need to add three electrons to the nearest vacant orbital of the neutral Cl atom. Now, the first vacancy is in the 3p orbital. Hence, the incoming electron will be added to 3p orbital.

Hence, the configuration of $P^{3-}$ ion is

p3– :12 252 2p6 3s2 3pნ

(d) $Si^{4-}$

Si has atomic number 14.

Hence, neutral Si atom has 14 electrons. Hence, the electronic configuration of neutral Si atom is

Si : 1s2 2s2 2p6 3s2 3p=

Now, to form $Si^{4-}$ ion, we need to add four electrons to the nearest vacant orbital of the neutral Si atom. Now, the first vacancy is in the 3p orbital. Hence, the incoming electrons will be added to 3p orbital.

Hence, the configuration of $Si^{4-}$ ion is

S4 : 1s2 2s2 2p6 3s2 3p-

Note that all the non-metal ions in this question has the same electronic configurations. Hence, they are all isoelectronic.

In fact all the ions we calculated the electronic configuration for has 18 electrons and are isoelectronic.