Arrange the following elements in the order of their increasing ionisation energies. O, S, Se, Te, Po
Se, Te, S, Po, O
O, S, Se, Te, Po
Po, Te, Se, S, O
Te, O, S, Po, Se
Arrange the following elements in the order of their increasing ionisation energies. O, S, Se, Te, Po
Se, Te, S, Po, O
O, S, Se, Te, Po
Po, Te, Se, S, O
Te, O, S, Po, Se
The correct option is C Po, Te, Se, S, O
Ionization enthalpy
The energy required to remove valence electron from an isolated gaseous atom is called ionization enthalpy.
Ionization enthalpy increases across a period from left to right due to the increase in the effective nuclear charge. The increase in nuclear charge makes the valence electron more bound to nucleus thus making it difficult to remove. Down a group, the increase in number of shells causes a decrease in effective nuclear charge on the valence electron thus decreasing ionization enthalpy.
The given elements belong to the 16th group.
Explanation for wrong answers
(A) Polonium (Po) is below Selenium (Se) and Tellurium (Te) in the 16th group. Since ionization enthalpy decreases down the group, the ionization energy of Polonium will be lower than Selenium and Tellurium. Option (A) is incorrect.
(B) Oxygen (O) is the first element in the 16th group. Therefore, it will have the highest ionization energy. The given order is the decreasing order of the ionization energies in the group. Therefore, option (B) is incorrect.
(D) The enthalpy of Oxygen will be highest and Polonium will be lowest in the 16th group. Therefore, option (D) is incorrect.
Explanation for correct answer
(C) The ionization enthalpy decreases down the group due to the increase in number of shells. The effect increase in the number of shells outweighs the effect in increase of electrons. Therefore, effective nuclear charge increases. The increasing order of atomic number of 16th group elements is O < S < Se < Te < Po.
Therefore, oxygen is the first element in 16th group and Polonium is the last. Since ionization energy decreases down the group, the increasing order of their ionization energies is Po < Te < Se < S < O.
The correct option is (C) Po < Te < Se < S < O.
Po, Te, Se, S, O
Ionization enthalpy
The energy required to remove valence electron from an isolated gaseous atom is called ionization enthalpy.
Ionization enthalpy increases across a period from left to right due to the increase in the effective nuclear charge. The increase in nuclear charge makes the valence electron more bound to nucleus thus making it difficult to remove. Down a group, the increase in number of shells causes a decrease in effective nuclear charge on the valence electron thus decreasing ionization enthalpy.
The given elements belong to the 16th group.
Explanation for wrong answers
(A) Polonium (Po) is below Selenium (Se) and Tellurium (Te) in the 16th group. Since ionization enthalpy decreases down the group, the ionization energy of Polonium will be lower than Selenium and Tellurium. Option (A) is incorrect.
(B) Oxygen (O) is the first element in the 16th group. Therefore, it will have the highest ionization energy. The given order is the decreasing order of the ionization energies in the group. Therefore, option (B) is incorrect.
(D) The enthalpy of Oxygen will be highest and Polonium will be lowest in the 16th group. Therefore, option (D) is incorrect.
Explanation for correct answer
(C) The ionization enthalpy decreases down the group due to the increase in number of shells. The effect increase in the number of shells outweighs the effect in increase of electrons. Therefore, effective nuclear charge increases. The increasing order of atomic number of 16th group elements is O < S < Se < Te < Po.
Therefore, oxygen is the first element in 16th group and Polonium is the last. Since ionization energy decreases down the group, the increasing order of their ionization energies is Po < Te < Se < S < O.
The correct option is (C) Po < Te < Se < S < O.
