Answer:
(a) 1s² 2s² 2p⁶ 3s² 3p⁴
(b) 1s² 2s² 2p⁶ 3s² 3p⁵
(c) sp³
(d) No valence orbital remains unhybridized.
Explanation:
<em>Consider the SCl₂ molecule. </em>
<em>(a) What is the electron configuration of an isolated S atom? </em>
S has 16 electrons. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁴.
<em>(b) What is the electron configuration of an isolated Cl atom? </em>
Cl has 17 electrons. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁵.
<em>(c) What hybrid orbitals should be constructed on the S atom to make the S-Cl bonds in SCl₂? </em>
SCl₂ has a tetrahedral electronic geometry. Therefore, the orbital 3s hybridizes with the 3 orbitals 3 p to form 4 hybrid orbital sp³.
<em>(d) What valence orbitals, if any, remain unhybridized on the S atom in SCl₂?</em>
No valence orbital remains unhybridized.
Answer:
An ionic bond forms between two ions of opposite charges. In ionic bonding, electrons transfer from one atom to another. The elements take on either a negative or positive charge. Ions are another name for charged atoms. Some elements are electropositive, and some are electronegative.
Hope this helps!
Answer:
Explanation:
Hello,
In this case, we can compute the required volume by using the ideal gas equation as shown below:
Thus, solving for the volume and considering absolute temperature (in Kelvins), we obtain:
Best regards.
Answer:
30 cm³
Explanation:
Step 1: Given data
- Density of aluminum (ρ): 2.7 g/cm³
- Mass of aluminum (m): 81 g
- Volume occupied by aluminum (V): ?
Step 2: Calculate the volume occupied by aluminum
The density of aluminum is equal to its mass divided by its volume.
ρ = m/V
V = m/ρ
V = 81 g / 2.7 g/cm³
V = 30 cm³
Answer:
54 days
Explanation:
We have to use the formula;
0.693/t1/2 =2.303/t log Ao/A
Where;
t1/2= half-life of phosphorus-32= 14.3 days
t= time taken for the activity to fall to 7.34% of its original value
Ao=initial activity of phosphorus-32
A= activity of phosphorus-32 after a time t
Note that;
A=0.0734Ao (the activity of the sample decreased to 7.34% of the activity of the original sample)
Substituting values;
0.693/14.3 = 2.303/t log Ao/0.0734Ao
0.693/14.3 = 2.303/t log 1/0.0734
0.693/14.3 = 2.6/t
0.048=2.6/t
t= 2.6/0.048
t= 54 days
