Halogens | Group 7 | Full Topic | Explained | A level Chemistry

[@umaraleem3476]

group 7 exam questions really have me in a twist, your video on that was very helpful! thank you, great prep for my mock on thursday.

[@chemistrytutor]

That's great to hear! Glad it's useful. Have you seen this one? kzbin.info/www/bejne/aJaxi6aBos2nrNE

[@MariamNuhu-fc9vz]

Another life saving video honestly making me so confident for my year 12 mock resits in 2 weeks, now all thats left are doing past paper questions Thanks sir for everything!!!

[@chemistrytutor]

The best of luck for your resits! 😀

[@Tharushi_SM]

13:34 stoichiometry of H20 is just 1. Not two here right? Thank you so much for this. I love how you connect all the parts to one another so it’s in a flow. Thank you again.

[@chemistrytutor]

Yes, great spot! I must have been thinking of the NaOH equation where you need the 2. Thanks for the feedback 😀

[@Tharushi_SM]

@@chemistrytutoryep, happy to help

[@sahib3035]

Oh my god I was looking for this the other day because I hate how group 7 exam equations are since I struggle to really understand halogens. words cant explain how happy i am that youve posted this THANK YOU!!

[@chemistrytutor]

Thank you so much for the kind words. This document of all my videos might be useful? drive.google.com/file/d/1s3I5prjbJRR1U1lcKpXO9TQpQMaVoP66/view?usp=drivesdk

[@Fall91894]

a really helpful video on this topic, it helped me a lot when going over my revision for the last time before my yr12 mock

[@chemistrytutor]

Excellent! I'm really glad it's useful 😊 Good luck for the mock 👍

[@RoseInnn]

How’d it go and how’s year 13 going for you?

[@TeekayOG]

this is some good stuff, really saving me a few days before my exam

[@chemistrytutor]

Excellent! Good luck with the preparation!

[@shahdamir4833]

BESTEST A LEVEL TEACHER

[@chemistrytutor]

Thank you 😊

[@nero3462]

Hi sir, you know when balancing the half equation at 26:43, how do we know the overall charge for each side? I hope this makes sense

[@chemistrytutor]

It makes perfect sense... you need to look at the charges on the ions- if there are any. The SO4 ion was 2- and each H+ was 1+ so 4H+ is 4+. The positives and negatives cancel down until only one type of charge is left, so it is 2+ overall since 4+ + 2- = 2+ Then on the right hand side we have two molecules, each with no charge, so the right hand side has no overall charge. This is why the electrons need to go onto the left hand side

[@andrewjustin256]

Hey there, I'm encountering difficulty with how you predicted what the products will be when NaOH reacted with Cl2 2NaOH + Cl2 --> NaCl + NaClO + H2O; how did you know it would yield sodium hypochlorite. If i were to be predicting on my own, i would simply think NaCl will be there only. Morever reaction like H2SO4 + NaCl --> HCl + NaHSO4. I thought the reaction would be like H2SO4 + 2NaCl --> 2HCl + Na2SO4 Please assist me in this!

[@chemistrytutor]

Great question! Predicting products can be tricky since it depends on reaction types and oxidation states. Let me break it down: Reaction 1: 2NaOH + Cl₂ → NaCl + NaClO + H₂O In this reaction, chlorine (Cl₂) undergoes disproportionation, meaning it’s both oxidized and reduced at the same time, forming products with two different oxidation states: 1. Recognizing Disproportionation: Chlorine starts in a zero oxidation state. When it reacts with a base (like NaOH), it splits into Cl⁻ (chloride) and ClO⁻ (hypochlorite) ions. 2. Final Products: - Chlorine is reduced to Cl⁻, which combines with Na⁺ to form NaCl. - Chlorine is also oxidized to ClO⁻, which pairs with Na⁺ to create NaClO. - Water (H₂O) is also formed as a byproduct. In alkaline conditions, chlorine often undergoes this kind of reaction, giving both NaCl and NaClO. Reaction 2: H₂SO₄ + NaCl → HCl + NaHSO₄ This reaction is controlled by the strength of sulfuric acid. Since it’s dilute here, it only acts as an acid, not an oxidizing agent. 1. Product Formation: H₂SO₄ reacts with NaCl, and the H⁺ from H₂SO₄ displaces Na⁺ from NaCl, creating HCl. 2. Why Not Na₂SO₄? Since only one mole of HCl is released per mole of NaCl, the leftover HSO₄⁻ ions pair with Na⁺ to form NaHSO₄. Forming Na₂SO₄ would need more NaCl and a more vigorous reaction. In summary: - For chlorine and bases, disproportionation is common, so look for products with different oxidation states. - For H₂SO₄ with salts like NaCl, expect NaHSO₄ when only single displacement occurs. Hope this helps!

[@andrewjustin256]

@chemistrytutor Thank you so much for replying me! May I also inquire from you whether there are more such rules or I should state "observations or good rule of thumb" as you have mentioned in the summary portion of your answer? Please refer to me a link or make a video on these seeming "exceptions" from the standard reactions (for instance Single or double displacement, redox, combustion, precipitation...) or "special conditions" of predicting the products! I also struggle with metalloids involvement in the reactions.

[@chemistrytutor]

Absolutely, I’m glad my response was helpful! You’re right-predicting products often goes beyond standard rules and benefits from knowing certain observations or "rules of thumb." Many reactions, especially in inorganic chemistry, have these “exceptions” that aren’t always intuitive at first but can be learned with experience. Here’s a bit more guidance: 1. Disproportionation Reactions: As with chlorine in NaOH, some elements (like halogens and transition metals) can simultaneously undergo oxidation and reduction. This tends to happen in reactions involving certain oxidizers or alkaline solutions. 2. Partial Displacement: When strong acids like H₂SO₄ react with salts (e.g., NaCl), partial displacement may produce NaHSO₄ instead of Na₂SO₄, especially when using a single mole of NaCl. Strong acid concentration, temperature, and other conditions often guide which product forms. 3. Common Reactions with Metalloids: Reactions involving metalloids like silicon or arsenic often follow specific patterns but can show unexpected results, particularly with strong acids or oxidizers. These elements may act as both metals and nonmetals depending on the context. For more on predicting products in these “special condition” reactions and exceptions, I recommend checking out this link on predicting chemical reactions which provides more details about the types of reactions, common exceptions, and the role of conditions in determining products. chemistrytalk.org/how-to-predict-chemical-reactions/ Thanks again for the thoughtful question-feel free to ask more if you come across them! 😊

[@armanazizi8817]

Great vid sirrrr

[@chemistrytutor]

Thank you for the feedback! 😊

[@Lily144atpase]

Could please make more video about inorganic chemistry?

[@chemistrytutor]

Yes, definitely. I've got all of Y12 done (other videos are Periodicity and Group 2). Y13 content will follow soon

[@Lily144atpase]

​@@chemistrytutor so reactions of cr and Mn in year 13? Sorry I'm just future international students and i want brush everything up before my bachelor degree. And in my country we cover an array of reaction and I'm a little bit lost

[@chemistrytutor]

@@Lily144atpase I'll definitely aim to cover some transition metals work for sure 👌

[@marlvrr]

thank you so much, great explanation, very helpful ❤

[@chemistrytutor]

I really appreciate the feedback! Thanks 😀

[@dsdsdzxc]

KING

[@chemistrytutor]

😃

[@zafranakram3098]

That’s sir , fantastic video very helpful ….. I’ll subscribe and watch your video s because you re very good at simplifying complex content for your students (like me) to understand !!

[@chemistrytutor]

Thank you, that's really kind and great to hear! This document of links will probably be useful! drive.google.com/file/d/1s3I5prjbJRR1U1lcKpXO9TQpQMaVoP66/view?usp=drivesdk

[@zafranakram3098]

@@chemistrytutor cheers !

[@minnaminnaminnaminna]

Hi sir, i dont fully understand why we acidify the silver nitrate with specifically nitric acid. I wonder if you could explain?

[@chemistrytutor]

Great question... There's three reasons 1) it already contains the nitrate ion that is the same as the second ion in silver nitrate. 2) it doesn't contain Cl- (and HCl does) 3) it doesn't contain a second ion that could get involved in redox. SO4^2- can be reduced by Iodide ions and so side reactions can occur

[@engineeringmadeasy]

Which exam board did you follow for A Level videos?

[@chemistrytutor]

Well, I teach AQA, so I make sure the videos cover everything for that exam board. Every topic for each exam board will be 98% identical. Its mostly the exam paper structure and question styles that vary 😀

[@TrishaLsg]

29:40 why do we add +8 charge on LHS. Where is the -8 so that it cancels out?

[@chemistrytutor]

We add the 10H+ to balance the H from the RHS and then we had 10 H+ and one SO4^2- so the overall charge was currently +8. Then we add the 8e- to bring the charge on LHS down to zero

[@TrishaLsg]

Rightt, in the overall equation. Thanks 🙏​@@chemistrytutor

[@gamesurfers2444]

Really helpful cheers.

[@chemistrytutor]

You're welcome! Thanks for the feedback 😀

[@intelblox7354]

At 9:44 are you doing these reactions in water?

[@chemistrytutor]

@intelblox7354 yes that's right. These are in aqueous solution (aq)

[@glo457]

23:42 im really confused, why do we get the same products in all three equations here? Why isn't So2 written as a product of the reaction with NaBr

[@chemistrytutor]

That section is to show you that they can react with the acid in a simple acid-base reaction producing similar products. On the next page, I then go on to beginning the redox interactions of Br^1- That's when SO2 is a product

[@glo457]

@@chemistrytutor does that mean the reaction of h2so4 with NaBr could either be redox or acid-base? do we need to know what causes the difference? and thankyou for the quick reply :)

[@chemistrytutor]

@glo457 in reality it's probably best to assume that both will happen simultaneously or even one after the other. Which is why in the redox equation you can see H+ and Br- on the left hand side... so effectively the HBr produced in the acid-base reaction can then reduce the S

[@sparkles6396]

hello, this was so helpful. btw sir, can i ask a question? so can you explain how to find angles when you are given a molceule such as ethanol or butanol? is there a easier way to find it...pls answer this, i have my exams next month

[@chemistrytutor]

I'm pleased it was useful. I'm not sure I know exactly what you mean. This is my video about shapes.... kzbin.info/www/bejne/iZy4noGCfLyGa9E Alternatively, you can think about the electron pairs around an atom. So the oxygen in ethanol has a bond to an H and one to the C. It also has 2 lone pairs. So 2 Bonding pairs and 2 lone pairs, means the bond angle between the 2 bonds will be the same as in water, so 104.5

[@sparkles6396]

@@chemistrytutor thank you very much sir, i will go thru the video as well.. 😁

[@zahramubashar7664]

Tyyyssssmmmm sir 😭

[@chemistrytutor]

Really happy to hear it made a difference-thanks!

[@minnaminnaminnaminna]

at 14:01 when you are writing Chlorine and water equations, wouldn't there be one mole of water as opposed to 2?

[@chemistrytutor]

Yes, great spot! I think I must have been thinking about the NaOH balancing which needs 2NaOH

[@Suadabdirizak]

Absolutely so beautiful

[@chemistrytutor]

😃

[@hasaanjunaid1842]

why should the metal halides be solid? doesnt (aq) work as well

[@chemistrytutor]

Which bit are you referring to?

[@hasaanjunaid1842]

@26:26 , does the reaction work with aq metal halides as well Pretty confusing, this.

[@hasaanjunaid1842]

Even at @24:09, all these metal halides. Can they be (aq)? If not then y? Thanks

[@chemistrytutor]

@hasaanjunaid1842 That's a great question! When reacting sodium or potassium halides with concentrated sulfuric acid, it's important to use the halide in its solid form rather than aqueous solution for a few reasons: 1. Chemical Reactivity: In their solid state, the halide ions are more reactive with concentrated sulfuric acid. This is because the reaction involves a redox process where the halide ions act as reducing agents. When the halides are in aqueous solution, the presence of water can interfere with the redox reaction, potentially diluting the acid and reducing its effectiveness as an oxidizing agent. 2. Gas Evolution: The reaction between solid halides and concentrated sulfuric acid typically results in the evolution of gases such as hydrogen halides (HX) or other gases depending on the halide used. If the halides were in solution, the evolved gases might dissolve back into the solution, which would not only reduce the yield but also make it difficult to collect and observe the gases. 3. Temperature Control: The reaction between solid halides and concentrated sulfuric acid is exothermic. Using a solid allows for better control over the reaction temperature. An aqueous solution could lead to a more rapid and less controlled reaction, increasing the risk of splattering and making it harder to manage. Additionally concentrated acids react very exothermically with water, and a small volume of solution in a test tube might not have enough water to dissipate all the heat energy. 4. Product Purity: Using solid halides helps in obtaining purer products. In an aqueous solution, side reactions can occur, leading to impurities in the final products.

[@hasaanjunaid1842]

Perfect Thanks

[@hasaanjunaid1842]

@24:19 cant the metal halides be (aq) Or will the reaction not work

[@chemistrytutor]

That's a great question! When reacting sodium or potassium halides with concentrated sulfuric acid, it's important to use the halide in its solid form rather than aqueous solution for a few reasons: 1. Chemical Reactivity: In their solid state, the halide ions are more reactive with concentrated sulfuric acid. This is because the reaction involves a redox process where the halide ions act as reducing agents. When the halides are in aqueous solution, the presence of water can interfere with the redox reaction, potentially diluting the acid and reducing its effectiveness as an oxidizing agent. 2. Gas Evolution: The reaction between solid halides and concentrated sulfuric acid typically results in the evolution of gases such as hydrogen halides (HX) or other gases depending on the halide used. If the halides were in solution, the evolved gases might dissolve back into the solution, which would not only reduce the yield but also make it difficult to collect and observe the gases. 3. Temperature Control: The reaction between solid halides and concentrated sulfuric acid is exothermic. Using a solid allows for better control over the reaction temperature. An aqueous solution could lead to a more rapid and less controlled reaction, increasing the risk of splattering and making it harder to manage. Additionally, concentrated acids react very exothermically with water, and a small volume of solution in a test tube might not have enough water to dissipate all the heat energy. 4. Product Purity: Using solid halides helps in obtaining purer products. In an aqueous solution, side reactions can occur, leading to impurities in the final products.

[@hasaanjunaid1842]

@@chemistrytutor perfect Thanks

[@hasaanjunaid1842]

Also Extremely sorry to disturb again and again. So what i understood was that, the reaction wont occur if metal halides were aq Could u then please help me with a past paper question about this? Its from chemistry AS level 9701, february march 2023, qp22, q2 d (i) In this strontium iodide (aq) is used along with concentrated sulfuric acid. But the reaction still occurs, according to the marking scheme. Shouldnt no reaction occur

[@chemistrytutor]

@@hasaanjunaid1842 😃

[@hasaanjunaid1842]

So i assume even if it’s aq, the reaction occurs. with that emoji😂😭😭

[@bentownsend4017]

🔥

[@chemistrytutor]

😀

[@inyourfacevia9708]

Why is it postive 2 charge instead of negative at 26:46

[@chemistrytutor]

That was a stepping stone calculation. RHS is 0 charge, the left *before electrons are added* The 4H+ ions and SO4^2- together have a 2+ charge which is why 2e needed to be added to make the charge zero

[@inyourfacevia9708]

@chemistrytutor i see!!! Thank you