(a)
Interpretation: The element having higher electronegativity is to be determined.
Concept introduction:
An atom's tendency to attract the bonding electrons into a covalent bond is known as electronegativity.
(a)
Answer to Problem 45A
F
Explanation of Solution
The electronegativity decreases down the group and increases across the period. Element Cl is placed below F. So, F will have higher electronegativity than Cl.
(b)
Interpretation: The element having higher electronegativity is to be determined.
Concept introduction:
An atom's tendency to attract the bonding electrons into a covalent bond is known as electronegativity.
(b)
Answer to Problem 45A
N
Explanation of Solution
The electronegativity decreases down the group and increases across the period. C is placed in group 14 and N is placed in group 15. So, nitrogen is placed right to the C.
Thus, the electronegativity of N is more than C.
(c)
Interpretation: The element having higher electronegativity is to be determined.
Concept introduction:
An atom's tendency to attract the bonding electrons into a covalent bond is known as electronegativity.
(c)
Answer to Problem 45A
Mg
Explanation of Solution
The electronegativity decreases down the group and increases across the period. Mg is placed below Ne, so Ne should have higher electronegativity than Mg. Since Ne is a noble gas, its electronegativity is zero.
Thus, among Mg and Ne, Mg will have higher electronegativity.
(d)
Interpretation: The element having higher electronegativity is to be determined.
Concept introduction:
An atom's tendency to attract the bonding electrons into a covalent bond is known as electronegativity.
(d)
Answer to Problem 45A
As
Explanation of Solution
The electronegativity decreases down the group and increases across the period. Ca is placed in group 2 and
Chapter 6 Solutions
Chemistry 2012 Student Edition (hard Cover) Grade 11
- The acid-base indicator HX undergoes the following reaction in a dilute aqueous solution: HX (color 1) H+ + X- (color 2). The following absorbance data were obtained for a 0.00035 M solution of HX in 0.1 M NaOH and 0.1 M HCI. Measurements were made at wavelengths of 450 nm and 620 nm using a 1.0 cm glass cuvette. 450 620 A(460 nm) A(630 nm) 0.1 M NaOH 0.1 M HCI 0.065 0.435 0.895 0.150 In the 0.1M NaOH solution, the indicator will be almost 100% in the X- form, while in 0.1M HCI, the indicator will be nearly 100% protonated (HX). Calculate the acid dissociation constant for the indicator if a pH=5 buffer solution containing a very small amount of indicator exhibits an absorbance of 0.567 at 450 nm and 0.395 at 620 nm (measured in a 1 cm glass cuvette).arrow_forwardShow work...give the name of the given compound. Don't give Ai generated solutionarrow_forwardShow work with explanation needed. don't give Ai generated solutionarrow_forward
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