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Class 11th The p-Block Elements

11.43. Assertion: Boron halides form addition compounds with NH₃.

Reason: Boron halides are Lewis bases.

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Vishal Baghel

Contributor-Level 10

(c) Boron halides are Lewis acids and can accept a pair of electrons from amines to form addition product.

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Class 11th The p-Block Elements

11.42. Assertion: Heavier elements form pπ– pπ bonds.

Reason: Their atomic orbitals are too large and diffuse to have effective overlapping.

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Vishal Baghel

Contributor-Level 10

(d) Heavier elementsdo not form pπ– pπ bonds because their atomicorbitals are too large and diffuse to haveeffective overlapping.

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Class 11th The p-Block Elements

Assertion and Reason:

Directions:

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

(c) Assertion is correct statement but reason is wrong statement.

(d) Assertion is wrong statement but reason is correct statement.

11.39. Assertion: Non-metals readily form anions, in contrast to metals whichreadily form cations.

Reason: In general, non-metals have higher ionisation enthalpies and higherelectronegativities than the metals.

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New answer posted

8 months ago

Class 11th Sets

43. Evaluate $\sum_{k = 1}^{11} (2 + 3^{x}) .$

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Payal Gupta

Contributor-Level 10

43. $\sum_{}^{11} (2 + 3^{x}) = (2 + 3^{1}) + (2 + 3^{2}) + . . . . . . . . . + (2 + 3^{11})$

$x = 1$ [2+2+ ….+ (11lines)] + (31+ 32+……+311)

= $11 * 2 + \frac{3 (3^{11} - 1)}{3 - 1} [? s_{n} = \frac{a (r^{n} - 1)}{r - 1}, r \geq 1]$

= $22 + \frac{3 (3^{11} - 1)}{2}$

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Class 11th Sequence and Series

41. 1, – a, a², – a³, . n terms (if a ¹ – 1).

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Payal Gupta

Contributor-Level 10

41. Here, $a_{1} = 1$

$r = \frac{- 9}{1} = - a$

So, $s_{n} = \frac{a_{1} [1 - r^{n}]}{1 - r}$

= $\frac{1 [1 - {(- a)}^{n}]}{1 - (- a)}$

= ${\frac{1 - (a)}{1 + a}}^{n}$

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Class 11th Sequence and Series

40. √7 , √21 , 3√7 , . n terms.

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8 months ago

Class 11th Sequence and Series

Find the sum to indicated number of terms in each of the geometric progressions inExercises 39 to 42:

39. 0.15, 0.015, 0.0015, . 20 terms.

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Payal Gupta

Contributor-Level 10

39. Here a = 0.15 = $\frac{15}{100}$

$r = \frac{0.015}{0.15} = \frac{\frac{15}{1000}}{\frac{15}{150}} = \frac{1}{10}$ <1

n = 20

So, Sum to term of G.P., $s_{n} = \frac{a (1 - r^{n})}{1 - r}$

$\Rightarrow$ $s_{20} = \frac{\frac{15}{100} [1 - {(\frac{1}{10})}^{20}]}{1 - \frac{1}{10}}$

= $\frac{\frac{15}{100} [1 - {(0.1)}^{20}]}{\frac{9}{10}}$

= $\frac{15}{100} * \frac{10}{9} [1 - {(0.1)}^{20}]$

= $\frac{1}{6} [1 - {(0.1)}^{20}]$

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8 months ago

Class 11th Sets

64. In a survey it was found that 21 people liked product A, 26 liked product B and29 liked product C. If 14 people liked products A and B, 12 people liked productsC and A, 14 people liked products B and C and 8 liked all the three products.Find how many liked product C only.

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Payal Gupta

Contributor-Level 10

64. Let A, B and C be the set of people who like product A, B and C respectively.

Then,

Number of people who like product A, n (A) = 21

Number of people who like product B, n (B) = 26

Number of people who like product C, n (C) = 29.

Number of people likes both product A and B, n (A $\cap$ B) = 14

Number of people likes both product A and C, n (A $\cap$ C) = 12

Number of people likes both product B and C, n (B $\cap$ C) = 14.

No. of people who likes all product, n (A $\cap$ B $\cap$ C) = 8

a→n (A $\cap$ B)

b→n (A $\cap$ C)

d→n (B $\cap$ C)

c→n (A $\cap$ B $\cap$ C)

From the above venn diagram we can see that,

Number of people who likes product C only

= n (C) - b - d + c

= n (C) - n (A $\cap$ C) - n (B $\cap$ C) + n (A $\cap$ B $\cap$ C)

= 29 - 12 - 14 +

...more

64. Let A, B and C be the set of people who like product A, B and C respectively.

Then,

Number of people who like product A, n (A) = 21

Number of people who like product B, n (B) = 26

Number of people who like product C, n (C) = 29.

Number of people likes both product A and B, n (A $\cap$ B) = 14

Number of people likes both product A and C, n (A $\cap$ C) = 12

Number of people likes both product B and C, n (B $\cap$ C) = 14.

No. of people who likes all product, n (A $\cap$ B $\cap$ C) = 8

a→n (A $\cap$ B)

b→n (A $\cap$ C)

d→n (B $\cap$ C)

c→n (A $\cap$ B $\cap$ C)

From the above venn diagram we can see that,

Number of people who likes product C only

= n (C) - b - d + c

= n (C) - n (A $\cap$ C) - n (B $\cap$ C) + n (A $\cap$ B $\cap$ C)

= 29 - 12 - 14 + 8

= 11

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New answer posted

8 months ago

Class 11th Sets

63. In a survey of 60 people, it was found that 25 people read newspaper H, 26 readnewspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T,8 read both T and I, 3 read all three newspapers. Find:

(i) The number of people who read at least one of the newspapers.

(ii) The number of people who read exactly one newspaper

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Payal Gupta

Contributor-Level 10

63. Let H, T and I be of people who reads newspaper H, T and I respectively.

Then,

number of people who reads newspaper H, n (H) = 25.

number of people who T, n (T) = 26.

number of people who I, n (I) = 26

number of people who both H and T, n (H $\cap$ I) = 9

number of people who both H and T, n (H $\cap$ T) = 11

number of people who both T and I, n (T $\cap$ I) = 8

number of people who reads all newspaper, n (H $\cap$ T $\cap$ I) = 3.

Total no. of people surveyed = 60

The given sets can be represented by venn diagram

(i) The number of people who reads at least one of the newspaper.

in (H∪T $\cup$ I) = n (H) + n (T) + n (I) n (H $\cap$ T) n (H $\cap$ I) n (T $\cap$ I) + n (H $\cap$ T $\cap$ I)

= 25 + 26 + 26 11 9 8 + 3

= 80 2

...more

63. Let H, T and I be of people who reads newspaper H, T and I respectively.

Then,

number of people who reads newspaper H, n (H) = 25.

number of people who T, n (T) = 26.

number of people who I, n (I) = 26

number of people who both H and T, n (H $\cap$ I) = 9

number of people who both H and T, n (H $\cap$ T) = 11

number of people who both T and I, n (T $\cap$ I) = 8

number of people who reads all newspaper, n (H $\cap$ T $\cap$ I) = 3.

Total no. of people surveyed = 60

The given sets can be represented by venn diagram

(i) The number of people who reads at least one of the newspaper.

in (H∪T $\cup$ I) = n (H) + n (T) + n (I) n (H $\cap$ T) n (H $\cap$ I) n (T $\cap$ I) + n (H $\cap$ T $\cap$ I)

= 25 + 26 + 26 11 9 8 + 3

= 80 28

= 52

(ii) From venn diagram,

a = number of people who reads newspapers H and T only.

b = number of people who reads newspapers H and I only.

c = number of people who reads newspapers T and I only.

d = number of people who reads all newspaper.

So, n (H $\cap$ T) = a + d.

n (H $\cap$ I) = b + d

n (T $\cap$ I) = c + d

So, a + d + c + d + b + d = n (H $\cap$ T) + n (H $\cap$ I) + n (T $\cap$ I)

a + d + c + b + 2d = 11 + 9 + 8

a + b + c +d = 28 - 2d

= 28 2 3 [d = n (H $\cap$ T $\cap$ I) = 3]

= 28 - 6

= 22

Number of people who reads exactly one newspaper

= Total no. of people - No. of people who reads more than one newspaper

= 52 - (a + b +c + d)

= 52 - 22

= 30

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8 months ago

Class 11th Sets

62. In a group of students, 100 students know Hindi, 50 know English and 25 knowboth. Each of the students knows either Hindi or English. How many studentsare there in the group?

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Payal Gupta

Contributor-Level 10

62. Let H and E be set of students who known Hindi and English respectively.

Then, number of students who know Hindi = n (H) = 100

Number of students who know English = n (E) = 50

Number of students who know both English & Hindi = 25 = n (H $\cap$ E)

As each of students knows either Hindi or English,

Total number of students in the group,

n (H $\cup$ E) = n (H) + n (E) - n (H $\cap$ E)

= 100 + 25

= 125,

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