Question

Which element in each of the following pairs has the larger ionisation energy (IE) ?
(a)$KorBr$
(b) $SorTe$
(c) $GaorSe$
(d) $NeorSr$

• A. $\left(a\right)Br\left(b\right)S\left(c\right)Se\left(d\right)Ne$
• B. $\left(a\right)Br\left(b\right)S\left(c\right)Ga\left(d\right)Ne$
• C. $\left(a\right)K\left(b\right)S\left(c\right)Ga\left(d\right)Sr$
• D. $\left(a\right)Br\left(b\right)S\left(c\right)Se\left(d\right)Sr$

Answer 1

The correct option is A $\left(a\right)Br\left(b\right)S\left(c\right)Se\left(d\right)Ne$

A quantitative measure of the tendency of an element to lose electron is given by its Ionization Enthalpy. It represents the energy required to remove an electron from an isolated gaseous atom (X) in its ground state. The first ionization enthalpy generally increases as we go across a period and decreases as we descend in a group.

a) In $K$ and $Br,K$ is metal and it's very easy for metal to loose an electron whereas it requires very high energy for a non-metal to loose an electron.

b) In $S$ and $Te,S$ requires more energy to loose an electron because ionisation enthalpy decreases as we descend in a group due to the decrease of the hold by nucleus on the outermost electron.

c) In $Ga$ and $Se,Se$ requires more energy more energy to loose an electron because ionization enthalpy increases as we go across a period.

d) In $Ne$ and $Sr,Ne$ requires much more energy because it is in most stable configuration and it will take very high energy to make $Ne$ lose an electron.