SCORING KEY



KENDRIYA VIDYALAYA SANGATHAN

HALF-YEARLY EXAMINATION- 2017

SET -2 scoring key

1. A REGENT WHICH first COMPLETELY CONSUMEd DURING THE REACTION.

2. Azimuthal or Angular momentum Quantum number.

3. NH3

4. ∆H = ∆E + ∆nRT ------------------- (½)

= 885389 + (-2)×8.314×300

= -880 KJmol -1 ------------------ (½)

5. I) HBr ----------------- (½)

ii) HClO4------ (½)

6. CH4(g)+ O2(g) → CO2(g) +2H2O(g) (½)

Correct answer 36 g of water (1 ½)

7. The atomic size ↓ as we move from left to right. Because within a period the outer electrons remain in the same shell but the nuclear charge increases for the addition of every electron which increases the nuclear attraction, thereby size decreases.(1)

In group, the atomic radius ↑ down the group. Because as we move down the e- s are added in the new shells, thereby the distance between the nucleus and outermost electron increases, which decreases the nuclear attraction. .(1)

8. a) Energy required to remove the most loosely bound electron from the neutral, isolated, gaseous atom. (1)

b) Energy involved during the addition of electron to the neutral, isolated, gaseous atom. (1)

(OR)

a) The physical and chemical properties of the elements are the periodic functions of their atomic numbers. (1)

b) Species which possess same no of electrons in their valence shell. (1)

9. P1V1 P2 V2

------ = -------- (½)

T1 T2

P2= 676.6 mm of Hg. (1 ½)

10. i)The solution which resists any change in its PH value even

When small amounts of the acid or the bases are added to it.(1)

Acid buffer : H2CO3 & NaHCO3 (1/2)

Basic buffer : NH4OH & NH4Cl (1/2)

11. i)One mole is the amount of substance that contains as many

particles or entities as there are atoms in exactly 12g of the 12C

isotope. ------------ ( 1 )

ii) A given compound always contains exactly the same proportion of elements by weight.…………….(1)

iii) Equal volume of all gases at the same temperature and pressure should contain same no of molecules.…………….(1)

12. (a) KE = ½ mv2 _____

v = √2KE = √2 x 3.0 x 10-25 kg m2s-2 = 812 ms-1 (1)

√m √9.1 x 10-3kg

λ = h = 6.626 x 10-34 kg m2s-1 (1)

mv (9.1x10-31kg) x 812 ms-1

λ = 8.967 x 10-7 m = 896.7nm ( 1 )

(OR)

a) ∆ x . m ∆ v = h/4 П ( 2 )

∆ v = h = 6.626 x 10-34 kg m2s-1

4П x ∆X x m 4x3.14 x 0.1 x 10-10 m x 9.11 x 10-31 kg

= 0.579 x 10 7 ms -1

∆ v = 5.79 x 10 6 ms -1 ( 1 )

13. a) i) Any two limitations (1)

b) i) In the ground state of the atoms, the orbitals are filed in order of their

increasing energies.…………….(1)

ii) Pairing of e’s in the orbitals belonging to the same sub shell does not

takes place until each orbital belonging to the same subshell has got one

electron each.……………. (1)

14. i) Due to the presence of same no of valence electrons.

ii) Na + &Mg 2+ (1)

K + Ca2+& S2- (1)

15. Correct MO Diagram with unpaired electron in outermost antibonding orbital.(2)

O22-< O2- ∆H, ∆G = -ve. ( 1 )

Hence the process is spontaneous.

ii) A real crystal has some disorder due to presence of defects, whereas ideal crystal has no disorder. ( 1 )

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