Periodic table Quiz-10

Periodic table Quiz-10

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Q1.   Molecular shape of SF_4,CF_4 and XeF_4 are:

•  The same with 2,0 and 1 lone pair of electrons respectively
•  The same with 1,1 and 1 lone pair of electrons respectively
•  Different with 0,1 and 2 lone pairs of electrons respectively
•  Different with 1,0 and 2 lone pairs of electrons respectively

Solution
(d) SF_4 has sp^3 d-hybridization with one lone pair; CF_4 has sp^3-hybridization with no lone pair and XeF_4 has sp^3 d^2-hybridization with two lone pairs.

Q2.   Which one of the following oxides is amphoteric in character?

•  SnO_2
•  SiO_2
•  CO_2
•  CaO

Solution
(a) A species is amphoteric if it is soluble in acid (behaves as a base) as well as in base (behaves as an acid.) SnO_2+4HCl→SnCl_4+2H_2 O basic acid SnO_2+2NaOH→Na_2 SnO_3+H_2 O acid base

Q3.   Length of hydrogen bond ranges from 2.5 Å to:

•  3.0 Å
•  2.75 Å
•  2.6 Å
•  3.2 Å

Solution
(b) H-bond has its bond length in the range 2.5 Å to 2.75 Å.

Q4.   o-hydroxy benzaldehyde, although contains enolic group but does not give test of group with FeCl_3 because:

•  It is steam volatile
•  Of intermolecular H-bonding
•  Of intermolecular H-bonding
•  All of the above

Solution
(c) Intramolecular H-bonding in salicyl aldehyde prevents its test with FeCl_3 (aq).

Q5.   Hydrogen fluoride is a liquid unlike other hydrogen halides because:

•  HF molecules associate due to hydrogen bonding
•  F_2 is highly reactive
•  HF is the weakest acid of all hydrogen halides
•  Fluorine atom is the smallest of all halogens

Solution
(a) A reason for given fact.

Q6.   Which of the following is correct order of increasing size?

•  Br^->S^(2-)>Cl^->Na^+>Mg^(2+)>Be^(2+)
•  Be^(2+)> Mg^(2+)> Na^+>S^(2-)>Cl^->Br^-
•  S^(2-)> Cl^->Br^->Na^+>Mg^(2+)>Be^(2+)
•  Na^+>Mg^(2+)>Be^(2+)>Br^->Cl^->S^(2-)

Solution
(a) Ions Be^(2+) Cl^- S^(2-) Na^+ Mg^(2+) Br^- Valence shell 1 3 3 2 2 4 Now, between Na^+ and Mg^(2+),Na^+>Mg^(2+) (isoelectronic), between Cl^- and S^(2-),S^(2-)>Cl^- (isoelectronic) because for isoelectronic species size decreases as the atomic number increases. Hence, the order of increasing size is Be^(2+)> Mg^(2+)> Na^+>Cl^->S^(2-)>Br^-

Q7.   Increasing order (lower first) of size of the various hybridised orbitals is:

•  sp,sp^2,sp^3
•  sp^3,sp^2,sp
•  sp^2,sp^3,sp
•  sp^2,sp,sp^3

Solution
(a) As the s-character increases in hybrid orbitals, bond energy increases, size of the hybridized orbital decreases. s-characters in sp,sp^2 and sp^3 are 1/2,1/3,1/4 respectively.

Q8.   Which is present in peroxides?

•  O_2
•  O^(2-)
•  O_2^(2-)
•  O_2^-

Solution
(c) [O-O]^(2-)

Q9.   Which of the following is the most electropositive element?

•  P
•  S
•  Mg
•  Al

Solution
(c) Electropositive character decreases across the period as metallic character decreases

Q10.   A molecule in which sp^2-hybrid orbitals are used by the central atom in forming covalent bond is:

•  He_2
•  SO_2
•  PCl_5
•   N_2

Solution
(b)