Question

1. Which of the following is a correct set of quantum numbers for an electron in a 4p orbital?

a. n = 4, l = 1, ml = 3

b. n = 4, l = 1, ml = +1

c. n = 4, l = 0, ml = -1

d. n = 4, l = 2, ml = 0

2. Arrange the elements K, Mg, Ar, and Na in increasing order of the energy required to remove the first electron from their respective gaseous atoms (IE1).

3. A.Circle the larger atom or ion out of each pair.

a. O or Se

b. Ca or Li

c. Cl or Cl

d. Ca2+ or Br27.

B. Circle the element out of each pair with the more negative electron affinity.

a. N or F

b. Na or S

Answer 1

1.

In 4p orbital

Principal quantum number (n) = 4

Azimuthal quantum number (l) = 1

Then, values of magnetic quantum number must be +1 , 0 or -1.

Hence, correct set is

n= 4, l= 1, ml = +1.

2. Noble gas Ar has complete valence shell (octet) hence Ar has the highest first ionization energy among given elements.

Electronic configuration of Mg is [Ne]3s2, hence 3s orbital is completely filled. So, its first ionisation energy will be more than Na (3s1) .

As , K has larger size than Na, less energy is required to remove the valence electron of K than that of Na.

Hence, increasing order of first ionization energy is

K < Na < Mg < Ar.

3.

Atomic size increases down a group and decreases across a period from left to right.

a) O, Se both are Group 16 elements, O is in period 2 , Se in period 4. Hence, Se is larger.

b)

Ca is larger ( period 4) than ( period 2).

c)

Anion is larger than neutral atom, Cl- is larger than Cl atom.

d) cation is smaller than anion.

Br- is larger than Ca2+  .

B.

a) Electrinic configuration of N is [He]2s22p3

Electronic configuration of F is [He]2s22p5 .

As, 2p orbital N is half filled , it is more stable than F(partially) .

Again after gaining an electron F atrain nibke gas lime electronic configuration. Therefore F has more negative electron affinity than that of N.

b)

Electron affinity increases across a period from left to right due to increase in nucleur charge .

Na and S belongs to same period . Na is in group 1, and S is in group 16, hence S has more negative electron affinity than Na.