Answer :
Let's break down the answers to these questions step by step:
### Question 10:
The effective charge [tex]\( Z^ \)[/tex] for a 3d electron in Selenium ([tex]\(_{34}\)[/tex]Se) is:
a) 21.15
b) 27.15
c) 15.15
d) 6.85
e) 12.85
Answer: c) 15.15
### Question 11:
Which element has the lowest first ionization energy?
a) Be
b) B
c) C
d) N
e) O
Answer: a) Be
### Question 12:
Arrange the following in order of decreasing ionic radius: [tex]\( P^{3-}, K^+, Ca^{2+} \)[/tex] and [tex]\( Cl^- \)[/tex].
a) [tex]\( P^{3-} > Cl^{-} > K^{+} > Ca^{2+} \)[/tex]
b) [tex]\( K^{+} > Cl^{-} > P^{3-} > Ca^{2+} \)[/tex]
c) [tex]\( K^{+} > P^{3-} > Cl^{-} > Ca^{2+} \)[/tex]
d) [tex]\( Ca^{2+} > Cl^{-} > P^{3-} > K^{+} \)[/tex]
e) [tex]\( Cl^{-} > P^{3-} > K^{+} > Ca^{2+} \)[/tex]
Answer: a) [tex]\( P^{3-} > Cl^{-} > K^{+} > Ca^{2+} \)[/tex]
### Detailed Explanation:
#### Question 10:
The effective nuclear charge [tex]\( Z^ \)[/tex] is an important concept in atomic chemistry. For a 3d electron in Selenium ([tex]\(_{34}\)[/tex]Se), the closest option to the true value of the effective nuclear charge is [tex]\( 15.15 \)[/tex].
#### Question 11:
Ionization energy refers to the energy required to remove an electron from an isolated atom. Among the given options, Beryllium (Be) has the lowest first ionization energy. This is because Beryllium has only two electrons in its outer shell, making it easier to remove one electron compared to the other elements.
#### Question 12:
The ionic radius decreases with increasing positive charge on the ion and decreases with decreasing negative charge. For the ions [tex]\( P^{3-} \)[/tex], [tex]\( Cl^- \)[/tex], [tex]\( K^+ \)[/tex], and [tex]\( Ca^{2+} \)[/tex]:
- [tex]\( P^{3-} \)[/tex] has the largest ionic radius because it has the highest negative charge.
- [tex]\( Cl^- \)[/tex] follows because it has a smaller negative charge.
- [tex]\( K^+ \)[/tex] is next, as it has a single positive charge, making it smaller than negative ions.
- [tex]\( Ca^{2+} \)[/tex] is the smallest because it has the highest positive charge.
Hence, the correct order in decreasing ionic radius is [tex]\( P^{3-} > Cl^- > K^+ > Ca^{2+} \)[/tex].
### Question 10:
The effective charge [tex]\( Z^ \)[/tex] for a 3d electron in Selenium ([tex]\(_{34}\)[/tex]Se) is:
a) 21.15
b) 27.15
c) 15.15
d) 6.85
e) 12.85
Answer: c) 15.15
### Question 11:
Which element has the lowest first ionization energy?
a) Be
b) B
c) C
d) N
e) O
Answer: a) Be
### Question 12:
Arrange the following in order of decreasing ionic radius: [tex]\( P^{3-}, K^+, Ca^{2+} \)[/tex] and [tex]\( Cl^- \)[/tex].
a) [tex]\( P^{3-} > Cl^{-} > K^{+} > Ca^{2+} \)[/tex]
b) [tex]\( K^{+} > Cl^{-} > P^{3-} > Ca^{2+} \)[/tex]
c) [tex]\( K^{+} > P^{3-} > Cl^{-} > Ca^{2+} \)[/tex]
d) [tex]\( Ca^{2+} > Cl^{-} > P^{3-} > K^{+} \)[/tex]
e) [tex]\( Cl^{-} > P^{3-} > K^{+} > Ca^{2+} \)[/tex]
Answer: a) [tex]\( P^{3-} > Cl^{-} > K^{+} > Ca^{2+} \)[/tex]
### Detailed Explanation:
#### Question 10:
The effective nuclear charge [tex]\( Z^ \)[/tex] is an important concept in atomic chemistry. For a 3d electron in Selenium ([tex]\(_{34}\)[/tex]Se), the closest option to the true value of the effective nuclear charge is [tex]\( 15.15 \)[/tex].
#### Question 11:
Ionization energy refers to the energy required to remove an electron from an isolated atom. Among the given options, Beryllium (Be) has the lowest first ionization energy. This is because Beryllium has only two electrons in its outer shell, making it easier to remove one electron compared to the other elements.
#### Question 12:
The ionic radius decreases with increasing positive charge on the ion and decreases with decreasing negative charge. For the ions [tex]\( P^{3-} \)[/tex], [tex]\( Cl^- \)[/tex], [tex]\( K^+ \)[/tex], and [tex]\( Ca^{2+} \)[/tex]:
- [tex]\( P^{3-} \)[/tex] has the largest ionic radius because it has the highest negative charge.
- [tex]\( Cl^- \)[/tex] follows because it has a smaller negative charge.
- [tex]\( K^+ \)[/tex] is next, as it has a single positive charge, making it smaller than negative ions.
- [tex]\( Ca^{2+} \)[/tex] is the smallest because it has the highest positive charge.
Hence, the correct order in decreasing ionic radius is [tex]\( P^{3-} > Cl^- > K^+ > Ca^{2+} \)[/tex].
