Four elements P- Q- R and S have ground state electronic configuration as-P 1 s - 2 - 2 s - 2 - 2 p - 6 - 3 s - 2 - 3 p - 3 1Q -rightarrow 1 s - 2 - 2 s - 2 - 2 p - 6 - 3 s - 2 - 3 p - 1 1R -rightarrow 1 s - 2 - 2 s - 2 - 2 p - 6 - 3 s - 2 - 3 p - 6 - 3 d 10- 4 s - 2 - 4 p - 3 S lrightarrow 1 s- 2 - 2 s - 2 - 2 p - 6 - 3 s - 2 - 3 p - 6 - 3 d -10- 4 s - 2 - 4 p - 1 Select the correct statement s among the following-A- Size of P - size of QB- Size of R - size of S-C- Size of P - size of R appreciable differenceD- Size of Q - size of S appreciable difference

The correct options are A Size of P lt; size of Q B Size of R lt; size of S C Size of P lt; size of R (appreciable difference)P Phosphorus (P), R A ...

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Byju's Answer

Standard XII

Chemistry

Multi-Electron Configurations and Orbital Diagrams

Four elements...

Question

Four elements P, Q, R and S have ground state electronic configuration as:
$P \to 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{3} Q \to 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1} R \to 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{3} S \to 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{1}$
Select the correct statement (s) among the following:

Size of

P

< size of

Q

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Size of

R

< size of

S

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Size of

P

< size of

R

(appreciable difference)

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Size of

Q

< size of

S

(appreciable difference)

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Solution

The correct options are
A Size of $P$ < size of $Q$
B Size of $R$ < size of $S$
C Size of $P$ < size of $R$ (appreciable difference)
P - Phosphorus (P), R - Arsenic (As)
Q - Aluminium (Al), S- Gallium (Ga)
As we know, according to the general trend, on going left to right in a period of the periodic table, size decreases and on moving top to bottom in a group, size increases.
But due to the poor screening effect of $3 d$ electrons, the size of $G a$ is less than that of $A l$
(A) $P < A l$
(B) $A s < G a$
(C) $P < A s$
(D) $A l > G a$

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Similar questions

Q. Group the following electronic configurations in pairs that would have different chemical properties of their atoms
(a)

{1 s}^{2}, {2 s}^{2}, {2 p}^{5}

(b)

{1 s}^{2}, {2 s}^{1}

(c)

{1 s}^{2}, {2 s}^{2}, {2 p}^{6}

(d)

{1 s}^{2}, {2 s}^{2}, {2 p}^{6}, {3 s}^{2}, {3 p}^{5}

(e)

{1 s}^{2}, {2 s}^{2}, {2 p}^{6}, {3 s}^{2}, {3 p}^{6}, {4 s}^{1}

(f)

{1 s}^{2}, {2 s}^{2}, {2 p}^{6}, {3 s}^{2} {3 p}^{6}, {4 s}^{2}, {3 d}^{10}, {4 p}^{6}

Q. Elements A, B, C, 0 and E having the following electronic configuration

(i)

1 s^{2}, 2 s^{2}, 2 p^{1}

(ii)

1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{1}

(iii)

1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{3}

(iv)

1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{5}

(v)

1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6}

Electronic configurations belonging to the same group are :

Q. The electronic configuration of the following elements K

^{19}

, Mn

^{25}

, Ca

^{20}

are given.

^{19}

= 1s

^{2}

^{2}

^{6}

^{2}

^{6}

^{x}

^{25}

= 1s

^{2}

^{2}

^{6}

^{2}

^{6}

^{5}

^{y}

^{20}

= 1s

^{2}

^{2}

^{6}

^{2}

^{6}

^{z}

The product of x, y, and z is:

Q. The excepted ground state configuration for

C u

is:

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{9}

The actual configuration is:

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1} 3 d^{10}

Why is the true electron configuration different?

Q. Sub-shell electronic configuration of elements P, Q and R is given.
[P, Q and R are imaginary symbols]
P

= 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}

,
Q

= 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}

,
R

= 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{2} 4 s^{2} 3 d^{1}

(a) One of these electronic configurations given above is wrong. Which is the element?
(b) Out of P, Q and R which has the highest electronegativity?
(c) Can P and Q join to form an ionic compound? Justify your answer.

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Four elements P, Q, R and S have ground state electronic configuration as: P→1s2 2s2 2p6 3s2 3p3Q→1s2 2s2 2p6 3s2 3p1R→1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3S→1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1 Select the correct statement (s) among the following: