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(i) What is meant by the term 'coordination number'? (ii) What
Chemistry Part-I
Chemistry Part-I

Class 12

NCERT

Publisher:
NCERT
Subject:
Chemistry
1

Chapter 1: The Solid State

31 questions

Q1
Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.
(i) Tetra phosphorus decoxide
(ii) Ammonium phosphate
(iii) SiC
(iv)
(v)
(vi) Plastic
(vii) Graphite
(viii) Brass
(ix) Rb
(x) LiBr
(xi) Si
Q2
(i) What is meant by the term 'coordination number'?
(ii) What is the coordination number of atoms:
(a) in a cubic close-packed structure?
(b) in a body-centred cubic structure?
Q3
How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.
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2

Chapter 2: Solutions

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3

Chapter 3: Electrochemistry

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4

Chapter 4: Chemical Kinetics

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5

Chapter 5: Surface Chemistry

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6

Chapter 6: General Principles and Processes of Isolation of Elements

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7

Chapter 7: The p-Block Elements

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8

Chapter 8: The d-and f-Block Elements

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9

Chapter 9: Coordination Compounds

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Question
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(i) What is meant by the term 'coordination number'?
(ii) What is the coordination number of atoms:
(a) in a cubic close-packed structure?
(b) in a body-centred cubic structure?

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Updated on: Apr 8, 2023

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Book:

Chemistry Part-I (NCERT)

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(i) What is meant by the term 'coordination number'?
(ii) What is the coordination number of atoms:
(a) in a cubic close-packed structure?
(b) in a body-centred cubic structure?
14 mins ago
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Text SolutionText solutionverified iconVerified

(i) Coordination number is the number of nearest neighbors with which a given atom is in contact. In an ionic crystal, the coordination number of an ion refers to the number of oppositely charged ions that surround that ion.
(ii) The coordination number of atoms in a:
(a) cubic close-packed structure is .
(b) body-centred cubic structure is .
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Practice more questions from Chemistry Part-I (NCERT)

Q1
Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.
(i) Tetra phosphorus decoxide
(ii) Ammonium phosphate
(iii) SiC
(iv)
(v)
(vi) Plastic
(vii) Graphite
(viii) Brass
(ix) Rb
(x) LiBr
(xi) Si
Q2
(i) What is meant by the term 'coordination number'?
(ii) What is the coordination number of atoms:
(a) in a cubic close-packed structure?
(b) in a body-centred cubic structure?
Q3
How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.
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Practice questions from Chemistry Part-I (NCERT)

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The correct statements(s) for cubic close packed (ccp) three dimensional structure is (are)
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The rate law and the rate constant from initial rate and concentration of the N O 2 - and O 3 Solution: R a t e = k N O 2 - 1 O 3 1 k = 5 × 10 5 M - 1 s - 1 Explanation: Concept: In this problem, we are given the initial rates of the experiments. So we can use the initial rate method to determine the order of N O 2 - and O 3 from the general rate law equation. Given: Experiment N O 2 - 0 (M) O 3 0 (M) Initial rate (M/s) 1 0.0100 0.0050 25 2 0.0150 0.0050 37.5 3 0.0200 0.0050 50.0 4 0.0200 0.0200 200.0 Formulae: The general rate law equation for the given reaction is R a t e = k N O 2 - m O 3 n where m and n are the order of N O 2 - and O 3 respectively. Calculation: The general rate law equation for the given reaction is R a t e = k N O 2 - m O 3 n To determine the value of n , we select the pair of experiment with the same concentration of N O 2 - , so they can cancel out. Experiment 3 and experiment 4 have the same value of N O 2 - . So we can take the ratio of rate 4 and rate 3 to calculate the order of O 3 . R a t e 4 R a t e 3 = k N O 2 - 4 m O 3 4 n k N O 2 - 3 m O 3 3 n 200.0 M / s 50.0 M / s = k 0.0200 M m 0.0200 M n k 0.0200 M m 0.0050 M n 4 = 0.0200 n 0.0050 n 4 = 4 n n = 1 So the order of O 3 is 1. Next, we select experiments 1 and 2, which have the same value of [ O 3 ] . So we can take the ratio of rate 2 and rate 1 to calculate the order m of N O 2 - . R a t e 2 R a t e 1 = k N O 2 - 2 m O 3 2 n k N O 2 - 1 m O 3 1 n 37.5 M / s 25 M / s = k 0.0150 M m 0.0050 M n k 0.0100 M m 0.0050 M n 1.5 = 0.0150 m 0.0100 m 1.5 = 1.5 m m = 1 So the order of N O 2 - is 1. Therefore the rate law for the given reaction is R a t e = k N O 2 - 1 O 3 1 = k N O 2 - O 3 Now we will calculate the k value from experiment 1. 25 M / s = k 0.0100 M 1 0.0050 M 1 k = 25 M / s 0.0100 M ( 0.0050 M ) 1 k = 5 × 10 5 M - 1 s - 1 Therefore, the rate constant is 5 × 10 5 M - 1 s - 1 Conclusion: We have calculated the rate law from the initial rate method and initial concentration given in the problem. We also calculated the rate constant from the rate law and initial concentration from experiments.
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Question Text
(i) What is meant by the term 'coordination number'?
(ii) What is the coordination number of atoms:
(a) in a cubic close-packed structure?
(b) in a body-centred cubic structure?
Updated OnApr 8, 2023
TopicThe Solid State
SubjectChemistry
ClassClass 12
Answer TypeText solution:1 Video solution: 1
Upvotes143
Avg. Video Duration2 min