Question

Multiple choice, choose the best correct answer: (50 pts, 1.5 for each) Which one of the species below should have the smallest value for tts firet fonieation en Rb Telo b) Na )INe Which of the following bonds exhibits the greatest ionic character b) H-I a) H-F c) H-Br In which structure(s) below does the oxygen have a formal charge of 17 H-ö-H H-01 H. Land IV and IV and 1 I only a) Il only b)

Answer 1

1)Option a is correct. Ionization energy is the amount of energy required to remove the electron from loosely bound outermost shell of neutral gaseous atom to convert it into positively charged ion. Larger the atomic size, less is the first ionization energy because outermost shell is at larger distance from nucleus, so electrons in outermost shell are loosely bound or experience less attraction from nucleus. Thus, the removal of electron in outermost shell is easy. is lNa and Rb belong to first group i. e. Alkali metals. Al belongs to group 13. Ne belongs to group 18 and O belong to group 16.O and Ne belong to same period i. e. Second period. O has 6electrons in valence shell whereas Ne has 8 electrons in valence shell. Ne has stable configuration as its octet is complete, so removal of electron from its outermost shell requires large amount of energy. However, o requires less energy in comparison to Ne to remove electron from its valence shell. For example: approximate value of first ionization energy of O is 1314KJ/mol and Ne is 2080KJ/mol.Na and Al belongs to same period i. e. Third period. On moving across a period from left to right atomic size decreases and removal of electron from loosely bound valence shell requires more energy. Na is on left side ( atomic number 11)and Al is on right side (atomic number 13).Thus,first ionization energy of Al is higher than Na because of small size. For example : first ionization energy of Na is 495KJ/mol and that of Aluminium is 776.8KJ/mol.

Rb is group one element which lies below Na. On moving down the group , size increases because of increase in number of valence shell. Therefore, Rb has largest size and less energy is required to remove electron from its outermost shell. Ionization energy of Na is 496KJ/mol and Rb is 403KJ/mol.

Moreover, alkali metals(Na and Rb) have largest size in comparison to other elements. Therefore, they have less ionization energy than other elements. Out of Na and Rb, Rb has large size so it has smallest ionization energy.

Increasing order of first ionization energy

Rb < Na<Al<O<Ne

2)Option a is correct. H-F exhibits greatest ionic character because of large electronegative difference. More the electronegativity difference between two elements, more is the ionic character. Here, electronegativity value of pauling scale are considered.

In case of H-F, electronegativity of F =4.0 and H = 2.1

electronegativity difference =4.0-2.1 = 1.9

In H-I, electronegativity of I = 2.5, electronegativity of H =2.1

Electronegativity difference = 2.5 - 2.1 =0.4

In H-Br, electronegativity of Br = 2.8, electronegativity of H =2.1

Electronegativity difference = 2.8-2.1 =0.7

In H-Cl electronegativity of Cl = 3.0 and electronegativity of H =2.1

electronegativity difference = 3.0 - 2.1=0.9

In H-H, both are same atoms

Electronegativity difference =2.1-2.1=0

The bond is purely covalent

Increasing order of ionic character

H-H< H-I< H-Br <H-Cl<H-F

3)Option e is correct.

Formal charge on O atom =[ number of electron in valence shell of O atom - ( 1/2 number of bonded electrons + number of unshared electrons )]

First structure  

Formal charge =[ 6 - (1/2 ×6 + 2)]

= 6 - (3+2) = 6-5 =+1

Second structure

Formal charge on O atom =[ 6-(1/2 ×2 + 6)]

= 6- ( 1+6) =6-7 =-1

Third structure

Formal charge on O atom = [6- (1/2 ×6+2)]

= [6-(3+2)]= 6-5 =+1

Fourth structure

Formal charge on O atom = [6-(1/2×6+2)]

= [ 6-(3+2)] =6-5 =+1