Question:

The electronic configuration of four atoms are given in brackets: $L\left(1s^{2} 2s^{2} 2p^{1}\right); $ $M\left(1s^{2} 2s^{2} 2p^{5}\right); $ $Q\left(1s^{2} 2s^{2} 2p^{6} 3s^{1}\right); $ $R\left(1s^{2} 2s^{2} 2p^{2}\right);$ The element that would most readily form a diatomic molecule is

Updated On: Jul 7, 2022
  • $Q$
  • $M$
  • $R$
  • $L$
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The Correct Option is B

Solution and Explanation

By sharing of $1$ electron each, both the atoms of $M$ will get a completed octet.
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Notes on Group 17 Elements

Concepts Used:

Group 17 Elements

Halogens are the group 17 elements of the periodic table. The term ‘halogen’ means ‘salt-producing’, hence the name halogens as they possess the tendency to form salts after reacting to metals. It generally has five elements:

  • Fluorine (F)
  • Chlorine (Cl)
  • Bromine (Br)
  • Iodine (I)
  • Astatine (At)

These are all naturally occurring halogens but Tennessine (Ts) is an artificially created halogen.

Halogens:

Halogens are highly reactive elements and are highly electronegative. They have a high tendency to react with metals to form salts. They are also known as Group 17 elements. They have 7 electrons in their outer shell with a configuration of (ns2 np5). Fluorine being the first halogen in group 17, is highly reactive. Astatine is a halogen because of its resemblance with iodine despite it being radioactive.

Electronic Configuration:

The general electronic configuration for group 17 elements is ns2np5. This configuration clearly shows that they have 7 electrons in their valence shell. They require one more electron to complete their octet and achieve noble gas configuration.