Question

Match the following

Column-I & Coloum-II \\\ \left( A \right)1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{1}} & \left( P \right)L\operatorname{argest}{{\left( I.E \right)}_{1}} \\\ \left( B \right)1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{5}} & \left( Q \right)L\operatorname{argest}{{\left( I.E \right)}_{4}} \\\ \left( C \right)1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{1}} & \left( R \right)L\operatorname{argest}{{\left( I.E \right)}_{3}} \\\ \left( D \right)1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}} & \left( S \right)L\operatorname{argest}{{\left( I.E \right)}_{1}} \\\ {} & \left( T \right)L\operatorname{argest}{{\left( I.E \right)}_{2}} \\\ \end{matrix}$$

Answer 1

As we know, ionization requires energy and the amount of energy which is required to separate one electron from its atom will depend on how tightly the electron is held. This is influenced by the number of protons and the orbitals in which the electron occupies.

Complete step by step solution:
- As we know, the protons in the nucleus are positively charged and electrons are negatively charged and there will be a force of attraction between them. The higher the pull of the nucleus, the tougher it will be to pull an electron away from an atom. The amount of energy needed to remove electrons from atoms can be termed as Ionization Energy.
- The first ionization energy is the energy to go from neutral atoms to cations with a +1 charge. Or in other words it is the energy needed to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous positive ions (cations).
- Also, the Atoms with more than one electron can have them successively removed and it will have successive ionization energy such as second ionization energy, third ionization energy and so on. The successive ionization energies will always be greater than the previous ionization energy. The reason for this property is that the electron is being pulled away from a more positive species.
- The atom which has the electronic configuration $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{1}}$(A) have the least number of electrons in the p orbital and if four electrons are removed, the proton to electron ratio will be highest in it making $L\operatorname{argest}{{\left( I.E \right)}_{4}}$.Thus A can be matched with Q.
- The atom which has the electronic configuration $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{5}}$ (B) have five electrons in the p orbital and it’s a configuration of a halogen and it will be highly difficult to remove the electrons making it $L\operatorname{argest}{{\left( I.E \right)}_{2}}$.Thus B can be matched with T.
- The atom which has the electronic configuration $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{1}}$ (C) have one electron in the s orbital and therefore it’s partially filled. It will be difficult to remove the electrons making it $L\operatorname{argest}{{\left( I.E \right)}_{3}}$.Thus C can be matched with R.
- The atom which has the electronic configuration $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}$ (D) is a fully filled orbital (octet configuration or noble gas configuration). Therefore it will be extremely difficult to remove the electrons making it $L\operatorname{argest}{{\left( I.E \right)}_{1}}$.Thus D can be matched with P.

Note: The difference between the terms ionization potential and ionization energy should be noted. Ionization potential is the potential needed to eject an electron from an atom and is usually expressed in electron volt whereas ionization energy or ionization enthalpy is the energy required to eject an electron from an atom which is usually expressed in Kilojoule per mole.