Which of the following is the CORRECT decreas | vedclass

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(A) Ionization enthalpy is the energy required to remove the most loosely bound electron from an isolated gaseous atom. As we move down a group in the

Vedclass · Accepted Answer

$S > Se > Te > Po$

Vedclass · Answer

(A) Ionization enthalpy is the energy required to remove the most loosely bound electron from an isolated gaseous atom. As we move down a group in the periodic table,the atomic size increases due to the addition of new shells. This increase in atomic size results in a decrease in the effective nuclear attraction on the valence electrons,making it easier to remove an electron. Therefore,the ionization enthalpy decreases down the group. For Group $16$ elements $(O, S, Se, Te, Po)$,the decreasing order of ionization enthalpy is $S > Se > Te > Po$.

Which of the following is the $CORRECT$ decreasing order of ionization enthalpies of the elements in Group $16$?

Similar Questions

Consider the following statements in respect of oxides of sulphur.
$(1)$ In gas phase $SO_2$ molecule is $V$-shaped.
$(2)$ In gas phase $SO_3$ molecule is planar.
$(3)$ $\gamma-SO_3$ is a cyclic trimer.
Which of the above statements are correct?

Which of the following statements is not correct?

Assertion : Both rhombic and monoclinic sulphur exist as $S_8$ but oxygen exists as $O_2$.
Reason : Oxygen forms $p\pi-p\pi$ multiple bond due to small size and small bond length but $p\pi-p\pi$ bonding is not possible in sulphur.

Choose the $CORRECT$ statement.

Identify the incorrect statement regarding ozone $(O_3)$.

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