Question

the total number of lone pairs in ncl3 is ____. 6 8 9 10 13 question 12 the lewis structure for a chlorate ion, clo3-, should show single bond(s), double bond(s), and __ lone pair(s). 2, 1, 10 3, 0, 9 2, 1, 8 3, 0, 10 2, 1, 9

Explanation:

Step1: Determine valence - electrons of N and Cl

N has 5 valence electrons and each Cl has 7 valence electrons. In $NCl_3$, the total number of valence electrons is $5 + 3\times7=26$.

Step2: Draw the Lewis - structure

N is the central atom. N forms 3 single bonds with 3 Cl atoms, using 6 electrons for bonding. So, $26 - 6 = 20$ electrons are left. Each Cl atom has 3 lone - pairs (3×3 = 9 lone - pairs on Cl atoms) and N has 1 lone - pair. The total number of lone - pairs is $9 + 1=10$.

for Question 12:

Step1: Determine valence - electrons of Cl and O in $ClO_3^-$

Cl has 7 valence electrons, each O has 6 valence electrons, and there is an extra electron due to the negative charge. So, the total number of valence electrons is $7+3\times6 + 1=26$.

Step2: Draw the Lewis - structure

Cl is the central atom. Cl forms 3 single bonds with 3 O atoms, using 6 electrons for bonding. So, $26 - 6 = 20$ electrons are left. One of the O atoms can form a double - bond with Cl to minimize formal charges. After distributing the remaining electrons as lone - pairs:

• There are 3 single bonds.
• There is 1 double bond.
• The number of lone - pairs: 10 (3 lone - pairs on each of the 2 single - bonded O atoms and 2 lone - pairs on the double - bonded O atom and 1 lone - pair on Cl).

Answer:

10