The correct statement among the following is
(a) (SiH3)3N is planar and less basic than (CH3)3N.
(b)

Q: Why doesn't Carbon form back bonds with Nitrogen?

Carbon is a second-period element. Its valence shell consists only of $2s$ and $2p$ orbitals, which are fully utilized in bonding. It lacks the empty $d$-orbitals required to accept electron density from Nitrogen's lone pair.

Q: How does back bonding change the shape from pyramidal to planar?

For effective $\pi$-overlap, the lone pair must reside in a pure $p$-orbital rather than a hybridized $sp^3$ orbital. This forces the Nitrogen atom to rehybridize to $sp^2$, which naturally adopts a trigonal planar geometry with $120^\circ$ bond angles.

Q: Does back bonding always decrease basicity?

Yes, if the lone pair responsible for basicity is delocalized into the back bond, its electron density is spread away from the central atom. This makes it less available to attract and bond with an incoming proton ($H^+$).

Q: Is the $p\pi-d\pi$ bond a true double bond?

It has partial double bond character. While not as strong as a standard $p\pi-p\pi$ double bond (like in $C=C$), it is significant enough to shorten the $N-Si$ bond length and enforce a planar geometry.

Q: Why is Silicon considered a 3rd-period element while Carbon is 2nd?

Silicon's outermost electrons are in the $n=3$ shell ($3s^2 3p^2$), which inherently includes the $3d$ subshell, even if it is empty. Carbon's outermost electrons are in the $n=2$ shell, which does not have a $d$ subshell.

Question

The correct statement among the following is

Options:

(A) $(SiH_3)_3N$ is planar and less basic than $(CH_3)_3N$.

(B) $(SiH_3)_3N$ is pyramidal and more basic than $(CH_3)_3N$.

(C) $(SiH_3)_3N$ is pyramidal and less basic than $(CH_3)_3N$.

(D) $(SiH_3)_3N$ is planar and more basic than $(CH_3)_3N$.

Sigma Percentile · Accepted Answer

### The Deceptive Similarity

At first glance, trimethylamine $(CH_3)_3N$ and trisilylamine $(SiH_3)_3N$ appear to be chemical twins. Both feature a central nitrogen atom bonded to three identical groups. You might intuitively expect them to share the same geometry and chemical properties. However, replacing carbon with silicon introduces a profound structural twist that completely alters the molecule's behavior.

### The Case of Trimethylamine

Let's start with the familiar territory of trimethylamine. The central nitrogen atom is bonded to three methyl groups and possesses one lone pair of electrons. Carbon, being a second-period element, has its valence shell strictly limited to $2s$ and $2p$ orbitals. It has no empty $d$-orbitals to offer.

Because there is nowhere else for the electrons to go, the lone pair remains strictly localized on the nitrogen atom. To minimize electron-electron repulsion, the nitrogen atom adopts an $sp^3$ hybridization. This gives the molecule a classic **pyramidal shape**, much like ammonia $(NH_3)$.

### The Twist with Trisilylamine

Now, enter trisilylamine. The nitrogen is now bonded to three silicon atoms. Silicon is a third-period element, which means its principal quantum number is $n=3$. This grants silicon access to the $3d$ subshell, which is entirely empty in its ground state.

This availability of empty $d$-orbitals changes the game. The lone pair on nitrogen, residing in a filled $2p$-orbital, senses the adjacent empty $3d$-orbitals of silicon.

### The Phenomenon of Back Bonding

Nature loves to delocalize electron density to achieve a lower energy state. The filled $2p$-orbital of nitrogen overlaps sideways with the empty $3d$-orbitals of the surrounding silicon atoms. This creates a **$p\pi-d\pi$ back bond**.

For this overlap to be geometrically possible and maximally efficient, the nitrogen atom must rehybridize. It shifts from $sp^3$ to $sp^2$ hybridization, placing the lone pair in a pure, unhybridized $p$-orbital perpendicular to the molecular plane. Consequently, the three $Si-N$ bonds spread out to $120^\circ$ angles, rendering the entire $(SiH_3)_3N$ molecule **perfectly planar**.

### The Basicity Showdown

Basicity in amines is defined by the availability of the nitrogen's lone pair to accept a proton $(H^+)$.

In trimethylamine, the lone pair is localized, concentrated, and highly available. It acts as a strong Lewis base.

In trisilylamine, however, the lone pair is heavily delocalized into the silicon atoms through back bonding. It is essentially "busy" holding the molecule together in its planar configuration. Because the electron density is spread out, it is far less available to attract a proton. Therefore, trisilylamine is significantly **less basic** than trimethylamine.

**Final Verdict:** $(SiH_3)_3N$ is planar and less basic than $(CH_3)_3N$.

Animated Solution for Chemistry - s and p-Block Elements: The correct statement among the following is

Select Answer:

Visualized Solution

$(C H_{3})_{3} N vs (S i H_{3})_{3} N$

\text{Lone Pair in } (CH_3)_3N

\text{Hybridization of } (CH_3)_3N

\text{The Case of } (SiH_3)_3N

p\pi-d\pi \text{ Back Bonding}

\text{Hybridization of } (SiH_3)_3N

\text{Comparing Basicity}

\text{Conclusion}

The Sigma Insight: Group 15 Elements

The Deceptive Similarity

The Case of Trimethylamine

The Twist with Trisilylamine

The Phenomenon of Back Bonding

The Basicity Showdown

Similar Questions

Based on the compounds of group 15 elements, the correct statement(s) is (are)

Which of the following statements is wrong?

Statement I: Sodium hydride can be used as an oxidising agent. Statement II: The lone pair of electrons on nitrogen in pyridine makes it basic. Choose the correct answer from the options given below.

In which of the following arrangements, the sequence is not strictly according to the property written against it?

Which of the following statements is true ?

White phosphorus on reaction with concentrated NaOH solution in an inert atmosphere of $CO_{2}$ gives phosphine and compound (X). (X) on acidification with HCl gives compound (Y). The basicity of compound (Y) is

On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with $H_{2}$ in the presence of a catalyst gives another gas (C) which is basic in nature. (A) should not be

Which of the following compound cannot act as a Lewis base?

Which one of the following group-15 hydride is the strongest reducing agent ?

Good reducing nature of $H_{3} P O_{2}$ is attributed to the presence of