Classification Of Elements & Periodicity In Properties MCQ - 1 (Advance) - JEE MCQ

Pick out the isoleelctronic structure from the following :

I. ⁺CH₃ 

II. H₃O⁺ 

III. NH₃ 

IV. CH₃^-

The size of the following species increases in the order :

Element in which maximum ionization energy of following electronic configuration would be

The outermost electronic configuration of most electronegative element is :

The electron affinity of the member of oxygen of the periodic table, follows the sequence

The process of requiring absorption of energy is

In the following which configuration of element has maximum electronegativity.

Highest size will be of

Atomic radii of flourine and neon in Å units are respectively given by

The correct order of second ionisation potential of C, N, O and F is :

Decreasing ionisation potential for K, Ca & Ba is

Element Hg has two oxidation states Hg⁺¹ & Hg⁺². the right order of radii of these ions.

The ionization energy will be maximum for the process.

Bond distance C - F in (CF₄) & Si - F in (SiF₄) are respective 1.33 Å & 1.54 Å. C - Si bond is 1.87 Å. Calculation the covalent radius of F atom ignoring the electronegativity differences.

Two elements A & B are such that B. E of A – A, B – B & A – B are respectively 81 Kcal/mole, 64 Kcal/mole, 76 Kcal/ mole & if electronegativity of B is 2.4. then the electronegativity of A may be approxiamtely

Question No. 20 and 21 are based on the following information.
Four elements, P,Q,R & S have ground state electronic configuration as :
P → 1s² 2s² 2p⁶ 3s² 3p³
Q → 1s² 2s² 2p⁶ 3s² 3p¹
R → 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p³
S → 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p¹

Q. Comment which of the following option represent the correct order of true (T) & false (F) statement.

I. size of P < size of Q

II. size of R < size of S

III. size of P < size of R (appreciable difference)

IV. size of Q < size of S (appreciable difference)

Four elements, P,Q,R & S have ground state electronic configuration as :
P → 1s² 2s² 2p⁶ 3s² 3p³
Q → 1s² 2s² 2p⁶ 3s² 3p¹
R → 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p³
S → 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p¹

Q. Order of IE₁ values among the following is