Question

(c) Why is Ce 3+ easily oxidised to Ce 4+?

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Answer 1

Answer:

-3,3,9,15,21

Your input -3,3,9,15,21 appears to be an arithmetic sequence

Find the difference between the members

a2-a1=3--3=6

a3-a2=9-3=6

a4-a3=15-9=6

a5-a4=21-15=6

The difference between every two adjacent members of the series is constant and equal to 6

General Form: a

n

=a

1

+(n-1)d

a

n

=-3+(n-1)6

a1=-3 (this is the 1st member)

an=21 (this is the last/nth member)

d=6 (this is the difference between consecutive members)

n=5 (this is the number of members)

Sum of finite series members

The sum of the members of a finite arithmetic progression is called an arithmetic series.

Using our example, consider the sum:

-3+3+9+15+21

This sum can be found quickly by taking the number n of terms being added (here 5), multiplying by the sum of the first and last number in the progression (here -3 + 21 = 18), and dividing by 2:

n(a1+an)

2

5(-3+21)

2

The sum of the 5 members of this series is 45

This series corresponds to the following straight line y=6x+-3

@Anirudh Mall

Answer 2

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One among the lanthanoids, Ce(III) can be easily oxidized to Ce(IV) (at. ... Ce(III) having the configuration 4f1 5d0 6s0 can easily lose electron to acquire the configuration 4f0 and form Ce(IV).

• brainlest Please ❤️✨