Question

The electrons in the atoms of two elements P and Q are distributed in three
shells having 1 and 7 electrons in their outermost shells.
i) Write the group numbers in which these elements are placed in the modern
periodic table.
ii) Write the electronic configuration of P and Q and iii)the molecular formula
of
the compound formed between P and Q. plz write correct

Answer 1

Answer:

i. Element P will be placed in the first group and Element Q will be placed in the 17th group of the Modern Periodic Table.

ii. Electronic Configuration of P = 2,8,1

Electronic Configuration of Q = 2,8,7

iii. The Molecular Formula of Compound of P and Q will be PQ.

I hope this helps.

Answer 2

P is in Group 1 and Q is in Group 17.

Their electronic configurations are  $1s^{2} 2s^22p^63s^1$ and $1s^{2} 2s^22p^63s^23p^5$.

The compound formed by them is NaCl.

Given:

Electrons in the third shell of P = 1

Electrons in the third shell of Q = 7

To Find:

Group numbers in which the elements are placed

Electronic configuration of P and Q

The molecular formula of the compound formed between P and Q

Solution:

Therefore as both P and Q have three shells, electrons in the first two shells = 2+8 = 10.

Electrons in the third shell of P = 1

Hence valence electrons = 1 and in s shell

Therefore P is placed in Group 1.

Q has 7 electrons in the third shell and therefore in the p shell.

Hence it is in Group 17.

The electronic configuration of P is $1s^{2} 2s^22p^63s^1$.

The electronic configuration of Q is $1s^{2} 2s^22p^63s^23p^5$.

Hence P is Na, and Q is Cl.

Therefore the formula of the compound formed between P and Q is NaCl.

#SPJ3