DPPs 4 - Career Point [PDF]

SO3 > SO2. Q.5. Which of the following species is/are having 'N–N' bond order = 2. (A) -. 3. N. (B) N2F2. (C) N2O4.

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CAREER POINT

Fresher Course for IIT JEE (Main & Advanced)–2017 DAILY PRACTICE PROBLEM SHEET Course : Fresher(XL) Batch Subject : Chemistry

DPPS-4

Topic : Chemical Bonding Q.1

Which of the following statements is/are true ? (A) Covalent bonds are directional (B) Ionic bonds are nondirectional (C) A polar covalent bond can be formed between two atoms of same electronegativity value (D) The presence of polar bonds in a polyatomic molecules suggests that it has zero dipole moment

Q.2

If IFxn , types species are planar and nonpolar, then which of the following match is correct (where x is number of F atoms and n is charge on species) (A) x = 2 and n = + 1 (B) x = 3 and n = 0 (C) x = 2 and n = – 1 (D) x = 5 and n = 0

Q.3

Choose the correct option for the following molecules in view of chemical bonding Cl H C=C=C=C H Cl (A) non-planar (B)  0 (C) A & B both (D)  = 0

Q.4

The correct order of 'S—O' bond length is(A) SO 32 > SO 24 > SO3 > SO2

(B) SO 32  SO 24 > SO2 > SO3

(C) SO 24  SO 32 > SO2 > SO3

(D) SO 24  SO 32 > SO3 > SO2

Q.5

Which of the following species is/are having 'N–N' bond order = 2 (A) N 3 (B) N2F2 (C) N2O4

(D) N2O

Q.6

Which of the following properties are same for N2H2, N2F2 and H2N2O2 compounds. (A) All are having two isomers (B) All are having equal number of covalent bonds between both N-atoms (C) All are having equal bond energy of N–N bond (D) Oxidation state of N-atom is same for all compounds.

Q.7

Which of the following statement is CORRECT(A) 'N—O' bond order is greater in 'NO' than that of NO2 (B) Both nitrous acid and nitric acid are planar species (C) 'S—O' bond length is greater in SO2 than that of SO3 (D) The Azide ion is isostructural with ozone molecule.

Q.8

Which of the following order is/are correct(A) CH4 > CH3F > CH2F2 > CHF3 : 'C – H' bonds length (B) CH3F < CH2F2 < CHF3

: 'C–F' bond length

(C) SeF4 < SeF3Cl < SeF2Cl2

: equatorial bond angle

(D) N2O4 > NO2 > NO 2

: O  N O bond angle



CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744 -5151200

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Page # 1

Q.9

The type of BACK BOND in compound (X) in following reaction is: ºC B2H6 + 2NH3  [BH2(NH3)2]+ [BH4]– 200   (X)

(A) (2p–2p) symmetrical distribution of electron density (B) (2p – 2p) unsymmetrical distribution of electron density (C) (2p – 3p) unsymmetrical distribution of electron density (D) (2p – 3p) symmetrical distribution of electron density Q.10

Q.11

Compounds are planar in its both monomer and dimer form. (A) 2NO2 Dimer  N2O4

(B) 2ICl3 Dimer  I2Cl6

(C) 2BeCl2 Dimer  Be2Cl4

(D) 2OF Dimer  O2F2

The CORRECT statement(s) regarding diborane (B2H6) is/are(A) Maximum six hydrogen atoms can lie in a plane (B) Maximum six atoms can lie in a plane (C) Bridging Hb – B – Hb bond is stronger than terminal B – Ht bond 



(D) Terminal H t  B H t bond angle is greater than bridging H b  B H b bond angle Q.12

Consider the following Ionisation processes NO  NO+ + e– N2  N 2 + e– O2  O 2 + e– Then according to the given information INCORRECT statement is: (A) Ionization energy : NO < O < N

(B) Ionization energy : O 2  O 2  O 2

(C) Ionization energy : N2 > O2

(D) NO+, N 2 , O 2 have fractional bond order

ANSWER KEY 1.(A,B,C)

2.(C)

3.(D)

4.(B)

5.(A,B)

8.(A,C)

9.(B)

10.(A,B,C)

11.(B,C,D)

12.(D)

CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744 -5151200

6.(A,B)

www.careerpoint.ac.in

7.(A,B,C,)

Page # 2

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