1s2 2s2 2p6 3s2 3p6 4s1
s orbital can hold 2 electron
p orbitals can hold 6 electron
Answer:
a) The lewis dot structure is shown in the image attached to this answer
b) The formal charge on each of the atoms is zero
c) bromine has an oxidation state of +5 while fluorine has an oxidation state of -1
d) 90 degrees
e) Square Pyramidal
f) polar bonds
g) polar molecule
Explanation:
The molecule BrF5 has a formal charge of zero. It exhibits an sp3d2 hybridization state with a square pyramidal geometry. The bond angle in the molecule is 90 degrees. It is a molecule of the type AX5E. The oxidation state of bromine is +5 while that of fluorine is -1.
The Br-F bonds are polar. The overall molecule is polar due to asymmetric charge distribution concentrating on the central atom since the molecule is square pyramidal.
1 mole of any particles = 6.02* 10²³ particles
4.5*10²⁵ atoms Ni* 1 mol Ni/6.02*10²³ Ni ≈ 74.75≈ 75 mol Ni
Answer:
what the heck what is that?
Explanation:
Answer:
Option B = 60,600 mg (correct option)
Explanation:
First of all we will have an idea which numbers are consider as significant.
1 = All non-zero digits are consider significant figures like 1, 2, 3, 4, 5, 6, 7, 8, 9.
2= Leading zeros are not consider as a significant figures. e.g. 0.02 in this number only one significant figure present which is 2.
3= Zero between the non zero digits are consider significant like 105 consist of three significant figures.
4= The zeros at the right side e.g 3400 are also significant. There are four significant figures are present.
In given options, Option A 60.6 mg have 3 significant figures.
Option B have 5 significant figures.
Option C have 4 significant figures.
Option D have 3 significant figures.
Thus option b is correct option which have more significant figures.
