MCQ
Electronic configuration of four elements $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and D are given below :
(A) $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{3}$
(B) $1 s^{2} 2 s^{2} 2 p^{4}$
(C) $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{5}$
(D) $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{2}$
Which of the following is the correct order of increasing electronegativity (Pauling's scale)?
  • A
    A $<$ D $<$ B $<$ C
  • B
    A $<$ C $<$ B $<$ D
  • C
    A $<$ B $<$ C $<$ D
  • D
    D $<$ A $<$ B $<$ C

Answer

D. D $<$ A $<$ B $<$ C
$\mathrm{N}:-1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{3}$ (Electronegativity $=3$ )
$\mathrm{O}:-1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{4}$ (Electronegativity $=3.5$ )
$\mathrm{F}:-1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{5}$ (Electronegativity $=4$ )
$\mathrm{C}:-1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{2}$ (Electronegativity $=2.55$ )
Correct order $=\mathrm{C}>\mathrm{B}>\mathrm{A}>\mathrm{D}$

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Product $(X)$ and $(Y)$ respectively is

Match the LIST-I with LIST-II.
LIST-I
(Family)
LIST-II
(Symbol of Element)
A.Pnicogen (group 15)I.Ts
B.ChalcogenII.Og
C.HalogenIII.Lv
D.Noble gasIV.Mc
Choose the correct answer from the options given below :