Answer:
Complete ionic:
.
Net ionic:
.
Explanation:
Start by identifying species that exist as ions. In general, such species include:
- Soluble salts.
- Strong acids and strong bases.
All four species in this particular question are salts. However, only three of them are generally soluble in water:
,
, and
. These three salts will exist as ions:
- Each
formula unit will exist as one
ion and one
ion. - Each
formula unit will exist as one
ion and two
ions (note the subscript in the formula
.) - Each
formula unit will exist as one
and two
ions.
On the other hand,
is generally insoluble in water. This salt will not form ions.
Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite
,
, and
(three soluble salts) as the corresponding ions.
Pay attention to the coefficient of each species. For example, indeed each
formula unit will exist as only one
ion and one
ion. However, because the coefficient of
in the original equation is two,
alone should correspond to two
ions and two
ions.
Do not rewrite the salt
because it is insoluble.
.
Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of
and two units of
. Doing so will give:
.
Simplify the coefficients:
.
Answer:
coordination number
Explanation:
Coordination number -
In a crystal lattice , the number of atoms that are surrounded to a particular atom , is referred to as the coordination number of the crystal.
In the field of crystallography and chemistry , it is also called the ligancy.
In coordination chemistry , the number of ligands attached to the central metal atom is also known as the coordination number of the coordination compound.
Hence, from the given statement of the question,
The correct term is coordination number.
Compound 1: Sodium borohydride
In sodium borohydride (NaBH4), B is a central metal.
Electronic configuration of B is 1s2 2s2 2p1.
B undergoes sp3 hybridization in NaBH4, to generate 4 hybrid orbitals. These hybrid orbitals, forms sigma bond with 4 'H' atoms. Due to this, the structure of sodium borohydride in tetrahedral.
........................................................................................................................
Compound 2: B<span>oron trifluoride
</span>In boron trifluoride (BF3), B is a central metal.
Electronic configuration of B is 1s2 2s2 2p1.
B undergoes sp2 hybridization in NaBH4, to generate 3 hybrid orbitals. These hybrid orbitals, forms sigma bond with 3 'H' atoms. Due to this, the structure of <span>boron trifluoride</span> is <span>triangular planner</span>.
This element is found in group 3A, period 3
<h3>Further explanation
</h3>
The maximum number of electrons that can be filled in the nth electron shell is 2n²(n=shell)
-
K shell (n = 1) maximum 2 x 1² = 2 electrons
- L shell (n = 2) maximum 2 x 2² = 8 electrons
- M shell (n = 3) maximum 2 x 3² = 18 electrons
- N shell (n = 4) maximum 2 x 4² = 32 electrons
Electron configuration of element X : 2.8.3 , so :
K shell = 2 ⇒1s²
L shell = 8⇒2s²2p⁶
M shell = 3⇒ 3s²3p¹
Block p: group 13-18 (has a 2p-6p configuration), also called a representative element because it includes metals, non-metals and metalloids
The outer shell 3s²3p¹ : located in group 3A and period 3
group⇒valence electron ⇒3
period⇒the greatest value of the quantum number n⇒3
CO isn’t a compound so it’s D